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DIFFERENTIAL EQUATIONS

(A Reference Text)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Prof. Francis Anthony G. Llacuna

Dr. Dante L. Silva

3rd  Quarter 

SY 2004-2005

 

 

 

 

 

 

 

TABLE OF CONTENTS

 

Lesson                                                                     Page

 

1          Classification of Differential Equations,

Its Definition and Terminology                                       4

 

2          Elimination of Arbitrary Constants by Differentiation                  11

 

3          Families of Curves                                                                                13

 

4          Variable Separable                                                                               14

 

5          Homogeneous Function                                                                        16

 

6          Equations with Homogeneous Coefficients                                             17

 

7          Exact Differential Equations                                                                   19

 

8          Reducing to Linear equation of Order One

                        Non-Exact Differential Equation                                                25

 

9          Transforming to Bernoulli’s Equation

                        Non-Exact Differential Equation                                                27

 

10        Integrating Factor Found by Inspection

                        Non-Exact Differential Equation                                                29

 

11        Integrating Factor Found by Formula

                        Non-Exact Differential Equation                                                31

 

12        Miscellaneous Substitution

                        Non-Exact Differential Equation                                                33

 

13        Coefficients Linear in Two Variables                                                     34

 

14        Geometric Application – Orthogonal Trajectories                                  36

 

15        Law of Growth and Decay                                                                    38

 

16        Newton’s Law of Cooling                                                                     40

 

17        Continuous Compound Interest                                                 42

 

18        Mixture Problem                                                                                   43

 

19        Simple Electric Circuits                                                             45

 

20        Newton’s Second Law of Motion                                                         47

 

21        Higher-Order, First Degree Differential Equations                                  49

 

22        Homogeneous Linear Equations with

                        Constant Coefficient                                                                 52

 

23        The Method of Undetermined Coefficients

                        Non-Homogeneous Equations                                      55

 

24        Variation of Parameters                                                                       

                        Non-Homogeneous Equations                                      58

 

25        R(x) is an Exponential Function

                        Inverse Operator                                                                      59

 

26        R(x) is a Trigonometric Function

                        Inverse Operator                                                                      61

 

27        R(x) is a Polynomial Function

                        Inverse Operator                                                                      63

 

28        R(x) is a Composite Function

                        Inverse Operator                                                                      64

 

29        By Inspection                                                                                       66

 

30        The Laplace Transform                                                             67

 

31        The Inverse Laplace Transform                                                 70

 

References                                                                                                       72

 

 

 

 

 

 

 

 

 

 

 

Lesson 1

 

Classification of Differential Equations,

 Its Definition and Terminology

 

Specific Objectives:

 

At the end of the lesson, the students are expected to accomplish the following:

·        define  differential equations

·        distinguish between dependent and independent variables

·        define parameter and arbitrary constants

·        classify differential equation according to properties, types, and kinds.

 

Differential Equation

A statement that two expressions are equal and an equation which contains derivatives or differentials. (Rainville, 2002)

 

 

Example 1.1 Which of the following describes a differential equation?

 

            1.  cos xdx + sin ydy = 0

           

            2.   – 3x + x2y = 0

 

            3. dy/dx = y/x + tan y/x

 

            4. xdx + ydy = 0

 

            5. (x2 – y2) + 2xy y’ = 0

 

Exercise 1.1 Which of the following describes a differential equation?

 

1.      (x + 3y)dx – (2x-y)dy = 0

 

2.      x2y’ + (y-1) = 2(x-3y)

 

3.      x2 – y2 = 9

 

4.      (x –y)dx/dy + 3x = x(x+y)

 

5.      sec x tan y – tan y = sec2 x – tan2 y

 

 

 

The Dependent and the Independent Variable

Suppose y = f(x), which is read as “y is a function of x”, if  y is differentiable with respect to x then y is the dependent variable while x is the independent variable.

 

Example 1.2 Identify the dependent and the independent variable.

           

            1. + k2x = 0                     t – independent variable; x – dependent variable

 

            2. = a2                    t, x – independent variable; w – dependent variable

 

Exercise 1.2 From the following differential equations, determine which among the         variables involved is the dependent variable and the independent variable:

     

      1. dy/dx + 1 = x – 2y

 

      2. (x-2y) 

      3. y’ – 2y= y” +2x

 

      4. (2x + y) dx + (x + 2y) dy = 0

 

      5. (cosx cosy) dx – (sinx siny) dy = 0

 

 

            Parameter is a constant not to be eliminated while an Arbitrary Constant is a constant to be eliminated.

 

Example 1.3 Identify the parameters.

 

            1.  5+ 6 + = 100w cos w:                w is a parameter/fixed constant

                                                                                     c is an arbitrary constant

 

Exercise 1.3 Identify the parameters.

           

            1. ( d2i / dt2) + di/dt + i/c = w coswt

 

 

 

 

 

 

Classification of Differential Equations

 

1. According to Properties: Order and Degree

 

Order of the Differential Equation

            The order of the differential equation is the order of the highest - ordered derivative appearing in the equation. (Rainville, 2002)

 

Example 1.4  Find the order of the following differential equations.

 

            1.  + k2x = 0          2nd order DE

 

Exercise 1.4  Find the order of the following differential equations.

 

1.      y”’ – 2(y”)2 = (y’–2y)3

 

2.     

 

3.      3

 

Degree of the Differential Equation:

The degree of the differential equation is the same as the exponent of the highest - ordered derivative in the given equation after the equation has been rationalized or cleared of fractions with respect to all the derivatives.

 

Example 1.5 Find the degree of the following differential equations

 

            1.  + k2x = 0       1st degree DE

 

           2.    

            Note; Square both sides of the equation to remove the                                                  fractional exponent, hence the degree of the equation is 2.

 

Exercise 1.5 Find the degree of the following differential equations.

 

  1. y”’ - 2(y”)3 = (y’)2 + 3y

 

 

 

2. According to Types: Ordinary and Partial DE

 

 

Ordinary Differential Equation

An equation which contains total derivatives or differentials only and has only one independent variable is called an ordinary differential equation.

 

Example 1.6 Which of the following describes an ordinary differential equations.

 

       1.  + k2x = 0:           t – independent variable; x – dependent variable

 

Exercise 1.6 Which of the following describes an ordinary differential equation?

 

  1. xy’ – y = 3(x+1)

 

  1. y’’-2y’+3y = 0

 

  1. x – 3y = 4

 

Partial Differential Equation

An equation which contains partial derivatives only and has one or more independent variables is called a partial differential equation.

 

 

Example 1.7 Which of the following describes a partial differential equation?

        

      1. Ά2w/Άt2   =  a2  (Ά2w/Άx2 )          t, x - independent variable ;  w - dependent variable

 

       2. cosx dy + siny dx = 0

 

Exercise 1.7 Which of the following describes a partial differential equation?

 

  1. ∂y/∂x – 3y∂y/∂z = (x- 2z)

 

  1. (x-2y)dx – (3x+y)dy = 0

 

  1. ∂f/∂t + ∂h/∂t = 0

 

3. According to Kind: Linear and Non-Linear

 

Linearity of an Equation

A differential equation is said to be linear if it satisfies the following conditions:

1. It should be linear in the dependent variable.

2. It should be linear in all the derivatives.

3. It should not contain a product of the dependent variable and its derivative or differential.

4. It should not contain any transcendental functions of the dependent variable

 

Note: If one of these conditions is not satisfied then the equation is considered non-linear.

 

 

Example 1.8 Determine which of the following differential equation is linear:

 

1.      (x + y - 2) dx – (x - 2y) dy = 0

 

2.      xdy/dx – 3y = 3x- 2x + 1

 

3.      (y’) - 2y = 3 (x+y)

 

4.      ydx/dy – 3x = x – 4y

 

5.      (y-1) dx – ( x+2) dy = 0

 

 

Exercise 1.8 Determine which of the following differential equation is linear:

 

            1. (y’’)2 + 10 (y’)3 + 3y = 3x4

 

            2. 7xy’’ + 13 x2y’ (cos x) y = 0

 

            3. (x – 2y)dy + (5x2 – 5y3)dx = 0

 

            4.  – 6– 5 = 0

 

            5. – 5– 15 = 0

 

Solution of a Differential Equation

The solution of a differential equation is a non-derivative relation between the variables which satisfies the given differential equation. It can be a General/Primitive Solution, Particular Solution or Singular Solution.

 

Exercise 1.9 Determine which of the following equations is a solution of the differential               equation:

 

  1. y = x - 2x + 3x – 4
  2. y’ = 3x - 4x + 3
  3. y = c1x -cx

 

 

1. General Solution or Primitive Solution

The general solution of a differential equation or primitive is a type of solution which contains arbitrary constant/s which is/are usually represented by letters. Geometrically, this represents family of curves.

 

Exercise 1.10 Determine which of the following equations is a general solution or           primitive:

 

1. y =

2. y = Ax + Bx + C

3. y = cx – ce

 

 

2. Particular Solution

The particular solution is obtained from the general solution by determining the value of the arbitrary constant and substituting it in the general solution.

 

Exercise 1.11 Determine which of the following equations is a particular solution.

 

  1. y = mx + b
  2. y =
  3. y = x - 2x + 2

 

3. Singular Solution

A solution free of arbitrary constants, and which is not obtained from the general solution.

 

Exercise 1.12

1.      Verify that y = e2x is a solution of

 

2.      Show that the differential equation (4x3 – y3)dx + (2y – 3xy2)dy = 0

            has x4 – xy3 + y2 = c as its solution where c is a constant.

 

Seatwork: Complete the following table:

 

 

Equation

 

Order

 

Degree

 

Kind

 

IV

 

DV

 

1. xdy + ydx = 0

 

 

 

 

 

2.

 

 

 

 

 

 

3.  y’’’ – 3y” + 3y’ + y = 0

 

 

 

 

 

4.

 

 

 

 

 

 

5. (y’’’)3 + (y”)4 + (y’)5 + y = 0

 

 

 

 

 

 

6. x (dy/dx) + x2 = 0

 

 

 

 

 

 

7. y (dy/dx) + x2 = 0

 

 

 

 

 

8.

 

 

 

 

 

 

9.

 

 

 

 

 

 

10. ∂y/∂x – 3y(∂y/∂z) = (x- 2z)

 

 

 

 

 

 

 

Homework: For each of the following differential equations, identify its order, degree,                            kind, type, independent and dependent variables. Present in tabular form.

 

            1. 5 (y”)2   +  10 (y’)3    +  3y  =   3x4

           

            2. 7xy”  +  13 x2y’ (cos x) y  =  0

           

            3. (x - 2y) dy  +   (5x2  - 5y3) dx =  0

           

            4. Ά2y/Άx2  -   6Άy/Άx  -  5  =  0

           

            5. (Ά3z/Άx3)2   -  (5Ά2z/Άy2)5  -  15Άz/Άt  =  0   

 

 

 

Review:  Differentiation Formula

 

Lesson 2

Elimination of Arbitrary Constants by Differentiation

 

Specific Objectives

            At the end of the lesson, the students are expected to:

·        be familiar with the method of eliminating the arbitrary constants from a given equation to obtain its differential equation  and

·        determine the differential equation from a given general solution or primitive.

 

Elimination of Arbitrary Constants:                                           

To find the differential equation from the given general solution, the arbitrary constants involved in the given equation must be eliminated.

 

Rules:

  1. Differentiate the given equation as many times as the number of arbitrary constants.
  2. The order of the resulting differential equation should be equal to the number of arbitrary constants.
  3. The desired equation is free from arbitrary constants.

 

 

Example: Find the differential equations of the following by eliminating the arbitrary                                 constants:

 

  1. y = 3x2 + ce-2x

 

  1. y = c1 sin4x + c2 cos4x + x

 

  1. y = (x + c) e-x

 

  1. x + y = tan (x + c)        Recall:  tan-1u =

 

  1. (x – a)2 + y2 = a2          Family of Circles with center at (a,0)

 

  1. x = b cos(wt + a)        Note:    Eliminate b and a, w is a parameter not to be                                                   eliminated, then differentiate with respect to t.

 

  1. y2 = cx                         Parabola with Vertex at (0,0) - (Board Problem)

 

  1. y = c1x + c2ex               (Board Problem)

 

  1. y = x + c1e-x + c2e-3x

 

 

Seatwork: Find the differential equations of the following by eliminating the arbitrary                               constants.

 

  1. y = 2x + cex                                         ans:  y’ -  y  +  2x  -  2  =  0

 

  1. y = c1ex + c2e2x + x                               ans:  y”  -  3y’  -   2x  +  2y  +  3  =  0    

 

 

Homework: Find the differential equations of the following by eliminating the arbitrary                             constants.

 

  1. y = c1e-2x + c2e3x                                   ans: y” – y’ – 6y = 0

 

  1. cxy + c2x + 4 = 0                                 ans: x3 (y’)2 + x2 yy’ + 4 = 0

 

  1. y = c1 ex cos3x + c2 ex sin3x                  ans: y” – 2y’ + 10y = 0

 

  1. ln y = c1x + c2e

 

  1. y = Ax + Bx + C

 

  1. y = c1 sin wt + c2 cos wt           w is a parameter

 

  1. (y – k)2 = 4a (x – h)                  parabola with vertex at (h,k), and a as parameter

 

  1. y = c1eaxcosbx + c2eaxsinbx       c1 and c2 are arbitrary constants and a and b are                                                            parameters

 

 

Review:  Equation of family of curves (straight line, parabola, circle, and ellipse)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 3

Families of Curves

 

Specific Objectives

            At the end of the lesson, the students are expected to:

·        formulate an equation based on the given condition

·        apply the appropriate method of eliminating the arbitrary constants.

 

A family of curves is usually defined by an equation containing one or more arbitrary constant and sometimes a parameter.

The Differential Equation of the family of the curves may be obtained by eliminating the arbitrary constants.

 

Example. Find the differential Equations of the following:

 

  1. Family of parabolas with vertex at (1,3) and axis parallel to the x–axis.                                                                                       ans.      2y’(x - 1) = (y – 3)

 

  1. All straight lines with slope and y-intercept equal.          

                                                                              ans.      y’(x + 1) – y = 0

                 

  1. Family of circles with center on the x–axis. (Board Problem)

                                                                                    ans.      1+yy” + (y’)2 = 0

 

  1. Family of circles through the origin with center on the x-axis.

                                                                                    ans.      2xyy’ + x2 – y2 = 0

 

  1. Family of circles with fixed radius r and tangent to the x-axis.

                                                                                    ans.      (y - r)2 (y’)2 + y2 – 2ry = 0

Seatwork:

 

  1. Find the DE in differential form the family of straight lines passing through the origin.       (Board Problem)                                  ans.      y dx – x dy = 0
  2. Find the DE in differential form the family of parabolas having vertices at the origin and then foci on the x axis.                ans.      y dx – 2xdy = 0

 

Homework:

 

  1. Find the DE of the family of circles with center on the y-axis.

                                                                              ans.      (y’)3 – xy” + y’ = 0

  1. Find the DE of the family of lines passing through (h,k).

                                                                                    ans.      (y-k) dx – (x-h)dy = 0

  1. Find the DE of the family of parabolas with axis parallel to the x-axis, the distance between the vertex and the focus being a.           

                                                                              ans.      2ay” + (y’)3  = 0

 

Lesson 4

Variable Separable

Solution of First Order, First Degree Differential Equations

 

 

Specific Objectives:

 

            At the end of the lesson, the students are expected to:

·        perform the solution to a differential equation and find the differential equations from a given relation

·        identify and distinguish differential equations whose coefficients are separable

 

Variable Separable Differential Equation

 

            Given the general form of differential equation,

 

                                    M(x,y)dx + N(x,y)dy = 0

 

            Suppose M(x,y)  and N(x,y) can be transformed to distinct functions of x and y as                                 M(x)N(y) and M(x)N(y) respectively, then the equivalent equation is

                                   

                                    M(x)N(y)dx + M(x)N(y)dy = 0

 

            Which can be transformed to:

                                     by means of separation of variables.

 

The resulting equation can now be integrated to find the general solution, thus

 

                                   

 

Note: To separate the variables, divide the equation by the unnecessary factor(s) or multiply the equation by the reciprocal of the unwanted factor(s).

 

 

Example: Obtain the general solution of the following differential equations.

 

  1. 3x2y dx – x3dy = 0                                            ans:      x3 = yc

 

  1. cosx siny dx + sinx cosy dy = 0     Recall:         

                                                                              ans:      sinx siny = c

 

  1. 2x(1-y2)dx – y(1+2x2)dy = 0                            ans:      (1 + 2x2)(1 - y2) = c

 

  1. (y2-1)dx – 2(2y + xy)dy = 0                             ans:      (x + 2) = c (y2 – 1)

 

  1. x siny dx + e-xdy = 0                                         ans:      xex – ex + ln(cscy – coty) = c

                  Recall the following:

1.      Integration By Parts (IBP)

2.     

                                               

  1. 2y2 dx + (x2y – 2xy) dy = 0                              ans:      y = cx / (x -2)  

                  Recall: Integration By Partial Fraction (IPF)

 

      7. y’ =                                                ans:      2ey =

 

 

Seatwork: Obtain the general solution of the following differential equations

 

  1. (xy2 + x) dx + (x2y – y) dy = 0                          ans: (x2-1)(y2 + 1) = c

 

  1. dx = t (1 + t2) sec2x dt                                                  ans: 2x + sin2x = (1+ t2)2 + c

 

 

Homework: Obtain the general solution of the following differential equations

 

     1. y’  =  y sec x  (Board Problem)                                        ans:  y  =  c(sec x  +  tan x)

 

     2. xy’ (2y - 1)  =  y (x  - 1)  (Board Problem)                      ans:  x  +  2y  +  c  = ln (xy)

 

     3. x (y  - 1) dx  +   (x  +  1) dy  =  0;  if  y  =  2

            when  x  =  1, determine y when x  =  2 (Board Problem)                                                                                                                                              ans:  y  =  1.55

 

    4. sin x sin y dx  +  cos x cos y dy  =  0                                 ans:  sin y  =  c cos x

 

    5. sec 2x cos y dx  +  (cos 2x  +  cos  2x cot2  y) dy  =  0 

                                                                                    ans:  tan 2x  +  2 ln ½csc y  -  cot y½  =  c

 

    6. (xy + x) dx = (x2y2 + x2 + y2 + 1) dy                     ans: ln ( x2 + 1)/(y +1)4 – y2 + 2y = c

 

    7. x (1-x2) dy = y dx                                                             ans: y2 (1- x2) cx2

 

 

Review: Homogeneous Function

 

 

Lesson 5

Homogeneous Function

 

 

The function f(x,y) = 0 is called a homogeneous function  of degree n if and only if   f(kx,ky) = kf(x,y), of degree n.

 

 

Example: Test whether the given function is homogeneous or not, if it is homogeneous                            then give the degree.

 

1.      f(x,y) = 4x2 – 3xy + y2

 

2.   f(x,y) = 2y +

 

            3.  f(x,y) = x sin (y/x) – y sin(x/y)         

4.  f(x,y) =

            5.  f(x,y) = x lnx – x lny

           

6.  f(x,y) = x3 – xy + y3

 

7.  f(x,y) = x lnx – y lny

 

 

Seatwork: Test whether the given function is homogeneous or not, if it is homogeneous                           then give the degree:

 

1.      f(x,y) = x + y - 2xy

2.      f(x,y) = ytan

 

Homework: Test whether the given function is homogeneous or not, if it is homogeneous                                    then give the degree:

1.      f(x,y) = 

2.      f(x,y) = x + y + sinx + cosy

 

3.      f(x,y) = xey/x

 

 

 

 

Lesson 6

Equations with Homogenous Coefficients

 

Theorem:

If M(x,y) and N(x,y) are homogeneous and of the same degree then

                         is homogeneous of degree zero.

 

If  M(x,y) and N(x,y) are homogeneous and of degree zero then the ratio                                             can be expressed as a function of a single variable                                                                     alone, say v.

 

Thus, the equation can be transformed to homogeneous type by replacing

                                    y = vx   dy = v dx + x dv

                        or         x = vy  dx = v dy + y dv

 

If both M and N are homogeneous and are of the same degree, then the differential equation M(x,y)dx + N(x,y)dy = 0 is homogeneous.

 

Note: Change the variable whose differential has simpler coefficients.

 

Example: Test whether each equation is homogeneous or not, if it is homogeneous then                          find the general solution.

 

1. + y)dx + xydy = 0                               ans: x2(x2 + 2y2) = c

 

2.  xy dx + 2 (x2 + 2y2) dy = 0              ans: (3x2 + 4y2)(y4) = c

 

3. ydx = (x + ) dy                             ans:

            Recall:

 

4. [ x csx (y/x) – y ]dx + x dy = 0                     ans: lnx – cos (y/x) = c

            Recall:

5. (x- xy  + y)dx – xydy = 0                        ans: c(y - x) =

 

            6.  x dx  +  sin2(y/x) [ y dx – x dy]  =  0            ans: lnx4 – 2(y/x)  +  sin2(y/x) =  c

                        Recall:

            7.  [ x – y tan-1(y/x)]dx + x tan-1(y/x) dy = 0      ans:

                        Recall: If we let u = ,   then du =

 

Seatwork: Test whether each equation is homogeneous or not, if it is homogeneous then                                     find the general solution.

 

 

            1. (x2  -  y2) dx  +  2xy dy  =  0                         ans:  x2  +  y2  =  cx

 

 

Homework: Test whether each equation is homogeneous or not, if it is homogeneous then                      find the general solution.

 

1.  xy(ydx +xdy) – 6ydy = 0 ;   when  y = 1 and x = 2            ans: y2x2 – 3y4 = 1

 

            2.    -  x dy  =  0 (Board Problem) 

                                                                                    ans:   +  y  =  c

            3.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 7

Exact Differential Equation

 

 

Specific Objectives

            At the end of the lesson, the students are expected to:

·        differentiate partially given functions of two variables

·        test given equations for exactness

·        solve equations using the four methods for exact equations

 

 

Consider the General Form of DE:

 

                                                M(x,y) dx + N(x,y) dy = 0

 

            Suppose separation of variables does not hold, assume that there is a function F(x,y) such that M =  and N = . Then differentiate M with respect to y and N with respect to x. That is,

 

                                                   

                                    but                  

Therefore,

                                                             

 

            which is a necessary and sufficient condition to be an exact equation.

 

Exact differential equations may be solved using any of the four methods:

  1. Integrable Combination
  2. Partial Derivatives
  3. Line integral
  4. Alternative Solution

 

 

1. Integrable Combinations

 

Integrable combinations consist of group of terms that forms an exact differential, thus it is readily integrable. It may be obtained by rearranging the terms in the given DE until a group of terms forms an integrable combination.

 

Some of the integrable combinations are listed below:

 

1.      xdy + ydx = d(xy)

2.     

3.     

4.     

5.     

6.     

7.     

8.     

9.      mxm – 1 yndx + nxm  yn – 1dy = d(xmyn)

10. mxm – 1 yndx – nxm  yn – 1dy/(yn)2 =  

 

Example 7-1: Find the general solution of the differential equation.

 

  1. (2x + 3y – 5) dx + (3x – y – 2) dy = 0           ans: 6xy + 2x2 – 10x – y2 – 4y = c

 

  1. (3x2y – 6x) dx + ( x3 + 2y) dy = 0                  ans: x3y – 3x2 + y2 = c

 

3. 3y (x2 – 1) dx + (x3 + 8y – 3x) dy = 0              ans: 4y2 + x3y – 3xy = c

 

4. (4x3y3 + 1/x) dx + (3x4y2 – 1/y) dy = 0                        ans: x4y3 + ln x/y = c

 

 

Seatwork 7-1: Find the general solution of the differential equation.

 

 

           1. +3)dx + (3y.      

                                                                                    ans:

       

           2.  2  

                                                                                    ans: 

 

Homework 7-1: Find the general solution of the differential equation.

 

1.      x (3xy – 4y3 + 6) dx + (x3 – 6x2y2 – 1) dy = 0

                                                           

                                                                        ans: x3y – 2x2y3 + 3x2 – y = c

 

2. (xy2 + x – 2y + 3) dx + x2y dy = 2(x+y)dy      when x = 1 and y = 1

                                                           

                                                            ans: (xy – 2)2 + (x+3)2 = 2y2 + 15

 

3. (2x + y + 1) dx + (x + 2y – 1) dy = 0        

                                                           

                                                            ans: x2  +  y2  +  x  -  y  +  xy  =  c

 

4. (cos x cos y – cot x) dx + (– sinx sin y) dy = 0   

                                                           

                                                            ans:  sin x cos y – ln sin x  =  c

 

            5. (2x3 – xy2 – 2y + 3) dx – (x2 y + 2x) dy = 0         

                                                                       

                                                                        ans:  x4  -  x2 y2  -  4xy  +  6x  =  c

 

2. Partial Derivatives

 

            Consider the equation: M(x,y) dx + N(x,y) dy = 0 ––––––––– (1)

 

            If , then (1) is an exact equation.

                                    Therefore, its solution is F = c where

 

–––– (2) and –––-– (3)

 

            Determine F from (2) by integrating both sides with respect to x, treating y as constant where the usual arbitrary constant in indefinite integration is a function T(y) which is yet unknown. To determine T(y), obtain, equate it to (3) and integrate. This time, no arbitrary constant is needed in obtaining T(y) since one is being introduced on the right side in the solution F = c.

 

 

Example 7-2: Find the general solution of the differential equation.

 

  1. (2x + 3y – 5) dx + (3x – y – 2) dy = 0           ans: 6xy + 2x2 – 10x – y2 – 4y = c

 

  1. (3x2y – 6x) dx + ( x3 + 2y) dy = 0                              ans: x3y – 3x2 + y2 = c

 

 

3. 3y (x2 – 1) dx + (x3 + 8y – 3x) dy = 0              ans: 4y2 + x3y – 3xy = c

 

4. (4x3y3 + 1/x) dx + (3x4y2 – 1/y) dy = 0                        ans: x4y3 + ln x/y = c

 

 

Seatwork 7-2: Find the general solution of the differential equation.

 

1. 3y (x2 – 1) dx + (x3 + 8y – 3x) dy = 0

 

2. (4x3y3 + 1/x) dx + (3x4y2 – 1/y) dy = 0                 

 

Homework 7-2: Find the general solution of the differential equation.

 

1.      x (3xy – 4y3 + 6) dx + (x3 – 6x2y2 – 1) dy = 0

 

2. (xy2 + x – 2y + 3) dx + x2y dy = 2(x+y)dy   when x = 1 and y = 1

 

3. (2x + y + 1) dx + (x + 2y – 1) dy = 0        

 

4. (cos x cos y – cot x) dx + (– sinx sin y) dy = 0   

 

5. (2x3 – xy2 – 2y + 3) dx – (x2 y + 2x) dy = 0         

 

 

3. Line Integrals

 

Consider      

 

 

                        where a, b = arbitrary constants for which M and N are defined.

 

            If M and N are polynomial functions, set a = b = 0 since a polynomial function is always defined at the origin except when the functions become undefined. In this case, try substituting other values for a and b that will make the functions defined.

 

Note:   In the first integral, treat y as constant.

In the second integral, replace x by the value of a.

 

            Applicable only when M and N are either polynomial functions or transcendental functions but never a combination of polynomial and transcendental functions.

 

 

Example 7-3: Find the general solution of the differential equation.

 

  1. (2x + 3y – 5) dx + (3x – y – 2) dy = 0           ans: 6xy + 2x2 – 10x – y2 – 4y = c

 

  1. (3x2y – 6x) dx + ( x3 + 2y) dy = 0                              ans: x3y – 3x2 + y2 = c

 

3. 3y (x2 – 1) dx + (x3 + 8y – 3x) dy = 0              ans: 4y2 + x3y – 3xy = c

 

4. (4x3y3 + 1/x) dx + (3x4y2 – 1/y) dy = 0                        ans: x4y3 + ln x/y = c

 

 

Seatwork 7-3: Find the general solution of the differential equation.

 

1. 3y (x2 – 1) dx + (x3 + 8y – 3x) dy = 0

 

2. (4x3y3 + 1/x) dx + (3x4y2 – 1/y) dy = 0                 

 

Homework 7-3: Find the general solution of the differential equation.

 

1.      x (3xy – 4y3 + 6) dx + (x3 – 6x2y2 – 1) dy = 0

 

2. (xy2 + x – 2y + 3) dx + x2y dy = 2(x+y)dy when x = 1 and y = 1

 

3. (2x + y + 1) dx + (x + 2y – 1) dy = 0        

 

4. (cos x cos y – cot x) dx + (– sinx sin y) dy = 0   

 

5. (2x3 – xy2 – 2y + 3) dx – (x2 y + 2x) dy = 0         

 

 

4. Alternative Solution

 

Consider the equation:    M(x,y) dx + N(x,y) dy = 0

 

            For      M: integrate x and treat y as a constant

                        N: set x = 0, then integrate y

 

 

Example 7-4: Find the general solution of the differential equation.

 

1.      (2x + 3y – 5) dx + (3x – y – 2) dy = 0           ans: 6xy + 2x2 – 10x – y2 – 4y = c

 

2.      (3x2y – 6x) dx + ( x3 + 2y) dy = 0                              ans: x3y – 3x2 + y2 = c

 

3. 3y (x2 – 1) dx + (x3 + 8y – 3x) dy = 0        ans: 4y2 + x3y – 3xy = c

 

4. (4x3y3 + 1/x) dx + (3x4y2 – 1/y) dy = 0      ans: x4y3 + ln x/y = c

 

 

Seatwork 7-4: Find the general solution of the differential equation.

 

1. 3y (x2 – 1) dx + (x3 + 8y – 3x) dy = 0

 

2. (4x3y3 + 1/x) dx + (3x4y2 – 1/y) dy = 0                 

 

Homework 7-4: Find the general solution of the differential equation.

 

1.      x (3xy – 4y3 + 6) dx + (x3 – 6x2y2 – 1) dy = 0

 

2. (xy2 + x – 2y + 3) dx + x2y dy = 2(x+y)dy     when x = 1 and y = 1

 

3. (2x + y + 1) dx + (x + 2y – 1) dy = 0        

 

4. (cos x cos y – cot x) dx + (– sinx sin y) dy = 0   

 

5. (2x3 – xy2 – 2y + 3) dx – (x2 y + 2x) dy = 0         

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 8

Reducing to Linear Equation of Order One

Non–Exact Differential Equations

 

 

Consider the General Form of DE

 

M(x,y) dx + N(x,y) dy = 0 ––––––––– (1)

 

                                                where .

 

            To transform (1) into an exact equation, multiply it by an integrating factor,(lambda).

 

Non - Exact differential equations may be solved using any of the four methods:

  1. Reducing it to Linear Equation of Order One
  2. Transforming to Bernoulli’s Equation
  3. Integrating Factor by Inspection
  4. Integrating Factor found by Formula (Partial Differentiation)

 

1. Reducing to Linear Equation of Order One

 

Specific Objectives

At the end of the lesson, the students should be able to :

·        identify a first order linear differential equation

·        find an integrating factor

·        solve a linear equation of order one

 

 

            If the given equation is not exact, reduce it to either form listed below and find an integrating factor,:

 

Linear in y: dy + y P(x) dx = Q(x) dx

G.S.:

 

Linear in x: dx + x P(y) dy = S(y) dy

 

Example: Find the general solution of the differential equation.

 

  1. (x5 + 3y) dx – x dy = 0                                    ans: 2y = x5 + cx3

 

2. dy = ( x- 2y cot 2x) dx                          ans: 4ysin2x = c–2x cos2x+sin 2x

                  Recall:

  1. 2y (y2 – x) dy = dx                                          ans:

 

  1. e-y sec2y dy = dx + x dy                                 ans: xey = tan y + c

                        Recall:

 

Seatwork: Find the general solution of the differential equation.

 

1.      dy + = x2dx                          ans: 4xy = x4 + c

 

2.      yey dx = ( y3 + 2x ey) dy                           ans: x + y2e-y = cy2

 

 

Homework: Find the general solution of the differential equation.

 

            1. (2x3 – y) dx + xdy = 0                                ans:  x3 +  y  =  cx

                                               

2. (2y sin x – cos3 x) dx + cos xdy = 0         ans:  y sec2  x  =  x  +  c

 

3. (3x2 + y + 3x3y) dx + xdy = 0                     ans:  e(xy  +  1)  =  c

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 9

Transforming to Bernoulli’s Equation

Non–Exact Differential Equations

 

 

2. Transforming to Bernoulli’s Equation

 

Specific Objectives

At the end of the lessons, the students should be able to :

 

            The standard form of the Bernoulli Equation takes a form similar to that of the linear differential equation of the first order. The main difference is on the factor yn and or xn contained at the right side of the Bernoulli equation:

 

Standard Forms:

 

dy + y P(x) dx = yn Q(x) dx  ––––––– (1)

 

dx + x P(y) dy = xn S(y) dy  ––––––– (2)

 

Consider (1) and multiply by y–n:

 

y–n dy + y1–n P(x) dx = Q(x) dx ––––––– (1’)

 

let u = y1–n

 

du = (1–n)y–n dy

 

Substitute in (1’):

 

                                      

 

 

  

 

 

 

 

Example: Find the general solution of the differential equation.

 

            1. dy + ydx = 2xy2ex dx                      ans:  1/yex  = - x2  +  c

 

            2.                 ans:  y  = x2 y sin y  +  x2 cos y  +  cx2                     

            3.   ans: (x2/y2)=  (-4/9) x3 -  (2/3) x3 ln x  +  c

           

            4. (3x2y2 – 3x2) dx + x3ydy – (y2 – 1) 3/2 dy = 0       

                                                            ans:  3x3 (y2 - 1)1/2  =  y3 -  3y  +  c

 

Seatwork: Find the general solution of the differential equation.

           

1. 2xyy’ = 4x2 + 3y2                                                    ans.  y2 = -4x2 + Cx3

 

            2. xydx + (x2 – 3y)dy = 0                                            ans.  x2y2 = 2y3 + C

 

            3.  2yxdx + x(x2lny – 1)dy = 0                                    ans. y (1 + x2-x2lny) = cx2

 

            4.  6y2dx – x(2x3 + y)dy = 0                                       ans. y = x3(2 + Cy1/2)

 

            5.  dy/dx = y(xy3 – 1)                                      ans. 3 = xy3 + y3 + Ce3xy3

 

Homework: Find the general solution of the differential equation.

 

1. x dy/dx + y = 1/y2                                                                    ans. y3 = 1 + cx-3         

 

2.   t 2dy/dt + y2 = ty                                                    ans.  et/y – ct

 

3.  2x3y’ = y(y2 + 3x2)                                                  ans. y2(c – x) = x3

 

4.  y’tanx sin2y = sin2x + cos2y                                 ans. (sin2x+ 3cos2y)sinx=c

 

5.   (y4 –2xy)dx + 3x2dy = 0 ,

when x =2, y = 1                                 ans. x2 = y3(x + 2)

 

 

Lesson 10

Integrating Factor Found by Inspection

Non–Exact Differential Equations

 

 

3. Integrating Factor Found by Inspection

 

Specific Objectives

            At the end of the lessons, the students should be able to :

·        identify the appropriate exact differentials in the equation

·        use the exact differentials in solving the given equation

·        solve a given differential equation observing the method

 

 

The achievement of a learner in finding the integrating factors by inspection depends largely upon his familiarity and creativity, but he might do well to keep in his mind the differentials of common functions, namely:

 

1.      xdy + ydx = d(xy)

 

2.     

3.     

4.     

5.     

6.     

7.     

8.     

9.      mxm – 1 yndx + nxm yn – 1dy = d (xmyn)

10. mxm – 1 yndx  – nxm yn – 1dy/(yn )2 =  

 

 

 

 

 

Example: Find the general solution of the differential equation.

 

1. 2xydx + (y2 – x2) dy = 0                             ans:  x2  +  y2  =  cy

 

2. 3x2y dx + (y4 – x3) dy = 0               ans:  3x3  +  y4  =  cy

 

3. ydx + (x + x3y2) dy = 0                               ans: -1/2(xy)2 + ln y  = c                    

 

            4 y(x2 + y) dx + x(x2 – 2y) dy = 0                   ans:  x2 y – y2  =  cx

           

 

Seatwork: Find the general solution of the differential equation.

 

1. (x2 + y2 – y) dx + xdy = 0                                       ans:

2. y(2xy + 1) dx – xdy = 0                                          ans: x2y + x = cy

 

3. y(x3 – y) dx – x(x3 + y) dy = 0                                ans:  x3 + 2y = cxy2  

 

 

Homework: Find the general solution of the differential equation.

         

1.   (x2y2 + 1)dx + x4y2dy = 0                         ans.  lnx3 + x3y3 = C

 

            2.  y(x3+x5)dx – x(x3+ y5)dy = 0                                 ans.  x4 – 4xy5 = Cy4

 

            3.   x4y’ = -x3y – csc(xy)                                             ans.  2x2cos(xy) – 1 = Cx2

 

           4    y(x3exy – y)dx + x(y + x3exy)dy = 0                        ans.  x3yexy  + y2 = Cx

 

           5.   y(y2 + 1) + x(y2 + 1)y’ = 0,

when x = 2 , y = -1                             ans. xy2 + x = -4y

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 11

Integrating Factor Found by Formula (Partial Differentiation)

Non–Exact Differential Equations

 

 

4. Integrating Factor found by Formula (Partial Differentiation)

 

Specific Objectives

At the end of the lessons, the students should be able to :

 

Consider the General Form of DE

 

M(x,y) dx + N(x,y) dy = 0 ––––––––– (1)

 

                                                where .

 

 

 

1.

 

 

2.

3.         If the given equation is a homogeneous equation, then

 

provided that xM + yN Ή 0.

 

4.         If the given equation can be transformed into the form

y f(x,y) dx  +  x g(x,y) dy = 0, then provided that xM – yN Ή 0.

 

Example: Find the general solution of the differential equation.

 

            1. (x2 + y2 + x) dx + xydy = 0             ans: 3x4  +  6 x2 y2 + 4x3 =  c

                                   

 

2. (x2  – y2) dx + xydy = 0                               ans:  2x2  ln x  +  y2  =  cx2

 

3. (xy2 + y) dx + (x2y – x) dy = 0                    ans:  xy  + ln (x/y)  =  c 

 

4. (2xy4ey + 2xy3 + y) dx + (x2y4ey – x2y2 – 3x) dy = 0        

                                                                        ans:  x2 y3 ey  +  x2 y2  +  x  =  cy3

 

 

Seatwork: Find the general solution of the differential equation.

 

            1.  y (4x + y) dx – 2(x2 – y)dy = 0                              ans. 2x2 + xy + 2ylnIyI = cy

 

            2.  y(y + 2x – 2)dx – 2(x + y)dy = 0               ans. x(2x + y) = cex

 

 

            3. (x2 + y2 + 1)  + x(x – 2y)y’ = 0                               ans. x2 – y2 + xy – 1= cx

 

 

Homework. Find the general solution of the differential equation.

 

            1.  y(8x – 9y)dx + 2x(x – 3y)dy = 0               ans. X3y (2x – 3y) = c

 

            2.  (2y2+3xy–2y +6x)dx + x(x +2y –1)dy = 0            ans. x2(y2 + xy –y + 2x) = c

 

 

More Exercises: Find the integrating factor of the following equations.

                       

            1.  2y(x2  –  y + x )dx + (x2 – 2y)dy = 0                     ans. e2x

 

            2.  y2dx + (3xy + y2 – 1)dy = 0                                   ans. y

 

            3.  y(2x – y + 1)dx + x(3x – 4y + 3)dy = 0                ans. y2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 12

Substitution Suggested by the Equation

(Miscellaneous Substitution)

Non–Exact Differential Equations

 

 

            In this topic, a new variable is introduced to simplify the form of the given equation and the choice depends on the form of the given equation. Usually, we replace the repeated quantity or the transcendental function by a new variable.

 

 

Example: Find the general solution of the differential equation.

 

            1. dx – (x + y)2dy = 0                                     ans: tan –1 (x + y) – y = c

 

            2.                                      ans: sin (y – x) = cx

 

            3.                                         ans:  xy2 – x2 = c

 

            4. (x2 + y2)(xdy + ydx) – xy(xdx + ydy) = 0    ans:  x2y =  c (x2  +  y2)

 

 

Seatwork: Find the general solution of the differential equation.

 

 

 

 

 

 

 

 

 

Homework: Find the general solution of the differential equation.

 

 

 

 

 

 

 

 

 

 

 

Lesson  13

Coefficients Linear in Two Variables

 

 

Consider the equation:

 

(a1x + b1y + c1) dx + (a2x + b2y + c2) dy = 0 –––––––– (1)

 

 

1. If c1 = c2 = 0, then (1) is a Homogeneous DE

 

2. If c1 and c2 are not both zero, then two cases arise, namely:

 

Case I: (intersecting lines)

 

To reduce (1) to a homogeneous DE, use the substitution

                           

x = u + h, dx = du  and  y = v + k, dy = dv –––––––– (2)

 

            where h, k are the coordinates of the intersection point.

– find h, k using the equations a1h + b1k + c1 = 0 and a2h + b2k + c2 = 0

– substitute h,k in (2) and in turn can be substituted in (1), making it a                                                                                                 homogeneous DE

 

            Substitute v = uz, dv = u dz  + z du

 

Case II. (parallel lines)

 

Use the substitution:

 

a1x + b1y = u              hence  a1 dx + b1 dy = du      or

a2x + b2y = u              hence  a2 dx + b2 dy = du

 

            which ever of the two has smaller coefficients and the resulting DE       becomes variable separable.

 

Example: Find the general solution of the differential equation.

 

            1. (x + y – 1)dx + (2x + 2y + 1)dy = 0           ans: x + 2y – 3ln ( x+ y+2) = c

 

            2. (2x + 3y – 5) dx + (3x – y – 2) dy = 0       ans: 2x2 - 10x + 6xy – 4y – y2 = c

 

            3. (6x – 3y + 2) dx – (2x – y – 1) dy = 0       ans: 3x – y – 5 ln (2x – y + 4) = c

 

 

 

Seatwork: Find the general solution of the differential equation.

           

 

            1. (2x + 3y – 1) dx – 4 (x + 1) dy = 0            ans: (2x – y + 3)4 = c (x + 1)3

 

 

Homework: Find the general solution of the differential equation.

 

            1. (2x + 3y – 1) dx + (2x + 3y + 2) dy = 0     ans: x + y + 3 ln (2x + 3y – 7) = c

           

            2. (x + 2y – 4) dx – (2x + y – 5) dy = 0         ans: (x – y – 1)3 = c (x + y – 3)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

          Lesson 14

Geometric Application – Orthogonal Trajectories

Elementary Applications of First–Order, First Degree

Ordinary Differential Equations

 

 

Specific Objectives

            At the end of the lesson, the student should be able to:

·        translate the rate of change into mathematical symbols,

·        apply the appropriate technique in solving a particular rate

equation,

·        use the concept of slopes to solve a geometric applications,

 

 

1. Geometric Application – Orthogonal Trajectories

 

            Two families of curves such that every member of either family cuts each member of the other family at right angles are called orthogonal trajectories.

 

 

where = slope of the original family of curves

 

         = slope of the other family of curves

 

 

Example: Find the orthogonal trajectories of the family of curves

 

                     1. y2 = 2x + c (Board Problem)                ans: y = ce –x

2. y = cx2                                                         ans:

3. x2 + y2 = cx                                     ans: x2 + y2 = cy

4. y2 = 4cx                                                      ans:

 

Seatwork: Find the orthogonal trajectories of the family of curves

 

1.      x – 3y = C.                                                ans. y + 3x = C

 

2.  Circles through the origin with centers on the x – axis.

Homework: Find the orthogonal trajectories of the family of curves

 

1. y ( x2 + k ) + 2 = 0                          ans.  y3 = -6 ln |cx |

 

2.      x = C exp (-y2)                                          ans.   y = C1 exp (x2)

 

3.      Ellipses with centers at (0,0) and two vertices at  (1,0) and (-1,0).

                                                                        ans.    x2 + y2 = 2 ln |cx|

 

4.      y = k ( csc x + cot x)                                 ans.    y2 = 2 (C – cos x )

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 15

Law of Growth and Decay

Elementary Applications of First–Order, First Degree

Ordinary Differential Equations

 

 

 

2. Law of Growth and Decay

 

            The rate of growth/decay of a certain population is directly proportional to the amount present.

                                               

                 –––– (1)                                           

                                   

                                    where = rate of growth/decay of a certain population

        k = constant of proportionality

              x = number of population present

                                           t = time

 

By separating the variables in (1), it yields to

 

Integrating both sides, we have

 

Example:

 

            1. Radium decomposes at a rate proportional to the amount at any instant. In 100 years, 100 mg of radium decompose to 96 mg. How many mg will be left after 200 years? (Board Problem)                           ans:     92.13 mg

 

            2. Initially, there are 250 bacteria and after seven hours, 800 bacteria are observed in the culture. Find:

a) the approximate number of bacteria that will be present in the culture after 24 hours

b) the time it will take the bacteria to increase to 2500.

 

                                                                                    ans:     a) x = 13,433 bacteria

                                                                                    b) t = 13.9 hrs.

            3. If the population of the country doubles in 50 years, how many years it would triple?                                                                                   ans:     79.2 yrs.

 

Seatwork:

           

            1. Initially, there is 50 mg of a certain radioactive material present and after 2 hours, it is observed that the material has lost 10 % of its original mass. Find:

a) the mass of the material after 4 hours

b) the time at which the mass has decayed to one–half of its initial mass (half–life of the material)

 

                                                                                                ans:     a) x = 40.45 mg

                                                                                                b) t = 13.10 hrs.

 

Homework:

 

            1. The population of a city doubles in 50 years. How many years will it be four times as much? Assume that the rate of change is proportional to the number of inhabitants.                                                       ans.  120 years

 

            2. A certain substance decomposes at a rate proportional to    the to the amount present. If one-third of it disappears after 1000 days, what is the percentage lost in 100 days?                                          ans.  3.97%

           

            3. Thirty percent of a radioactive substance disappears in 15   years.   Find the half – life of the substance.                             ans.  29.12 years

           

            4. Sugar decomposes in water at a rate proportional to the amount still unchanged. If there were 50 lb of sugar present initially and at the end of 5 hours this is reduced to 20 lb, how long will it take until 90% of the sugar is decomposed?

           

            5. Water leaks from a cylinder at a rate proportional to the square root of the volume remaining at any time. If initially there are 64 gallons present and 15 gallons leak out on the first day, how much will be left after 4 days? When will there be 25 gallons?                                                                 ans. 16 gal;  3 days

 

 

 

 

 

 

Lesson 16

Newton’s Law of Cooling

Elementary Applications of First–Order, First Degree

Ordinary Differential Equations

 

 

 

3. Newton’s Law of Cooling

 

 

            Newton’s Law of Cooling states that the time rate of change of the temperature of the body is proportional to the temperature difference between the body and its surrounding medium.

                                           

                                                           

                       

                                                            ––––- (1)

 

                        where =   time rate of change of the temperature of the body

                           T =    temperature of the body

                           Tm = temperature of the medium

                            k =   constant of proportionality

                                        t =    time

            By separating the variables in (1), it yields to

 

                                               

           

            Integrating both sides, we have

 

                       

 

 

Example: Solve the following

           

            1. A body at a temperature of 50°F is placed in an oven whose temperature is kept at 150° F. If after 10 minutes the temperature of the body is 75°F, find the time required the body to reach a temperature of 100°F

 

                                                                                                ans: t = 24.10 mins.

 

            2. A body with a temperature of 100°C is immersed in a liquid that is kept at a constant temperature of 20°C. After 15 minutes, the temperature of the body decreased to 60°C.

                        a) Find the time it will take for the temperature of the body to                                                     decrease to 40°C.

                        b) What will be the temperature of the body after 40 minutes?

 

                                                                                                ans:     a) t = 30 mins.

                                                                                                b) T = 32.6°C

 

 

Seatwork: Solve the following:

 

            1. If a thermometer is taken from a room in which the temperature is 75ΊC into the open, where the temperature is 35ΊC, and the thermometer reading is 65ΊC after 30 sec,

                        a) How long after the removal will the reading be 50ΊC?

                        b) What will be the thermometer reading 3 minutes after the                                                       removal?

                                                                                                ans.     a) 102 sec 

                                                                                                            b) 42.1°C

 

Homework: Solve the following

 

            1. At 1:00 pm, a thermometer reading 70Ί F is taken outside where the air temperature is - 10Ί F. At 1:02 pm, the reading is 26Ί F. at 1:05 pm, the thermometer is taken back indoors where the air is at 70Ί F. What is the thermometer reading at 1:09 pm?                              ans.  56° F

 

            2. A  body at a temperature of 50Ί F is placed in an oven whose  temperature is kept at 150Ί F. If after 10 minutes the temperature of the body is 75Ί F, find the time required for the body to attain a temperature that is within a half degree of the oven temperature.                                 ans. 184.16 min

 

 

 

 

 

 

 

 

 

 

Lesson 17

Continuous Compound Interest

Elementary Applications of First–Order, First Degree

Ordinary Differential Equations

 

 

4. Continuous Compound Interest

 

Rate Equation:          α P            or           =  r P

 

             Let       P = amount of money at any time

                        r = annual rate of interest

                        t = time

 

            By separating the variables, it yields to

 

           

            Integrating both sides, we have


 

 

Example: Solve the following:

 

            1. An annual interest of 4% is given on an account . What is the accumulated amount of Php 100,000 after 5 years?            ans.  Php 122,140.27

 

            2. If Php 50,000 is borrowed at  an interest of 12% per year  compounded continuously, and the loan is to be repaid in one payment at the end of 2 years,  how much must the borrower repay?                           ans. Php 63,562.05

           

 

Seatwork: Solve the following

 

            1. If an amount of money doubles itself in 10 years at an interest compounded continuously, how long will it take for the original amount to triple itself?                                                                                       ans.  15.85 years

 

 

 

 

Lesson 18

Mixture Problem

Elementary Applications of First–Order, First Degree

Ordinary Differential Equations

 

 

5. Mixture Problem

 

            The mixing of two salt solutions of differing concentrations gives rise to a first–order differential equation for the amount of salt contained in the mixture. If x(t) denotes the amount of salt in the tank at time t, then the rate at which x(t) changes is a net rate:

 

                                    where  and

            Let       x    = amount of substance present at any time

                        qi    = volumetric flow rate of the  solution coming in

                        qo  = volumetric flow rate of the solution going out                                                            ci   = concentration of the solution coming in

                        co  = concentration of the solution going out

                        t   = time

                         = rate of change

 

Example: Solve the following

 

            1. A tank initially contains 100 gal. of salt solution, where 50 lbs. of salt is added. Salt solution containing 1 lb/gal of salt goes into the tank at the rate of 2 gal/min and the solution thoroughly mixed goes out at the rate of 1 gal/min. Find the amount of pure salt after 100 minutes.                              ans: x = 175 lbs.

 

            2. Find the amount of pure salt after 100 minutes described in the previous problem if the solution thoroughly mixed out goes at the rate of 2 gal/min.          

                                                                                                            ans: x = 92.93 lbs.

 

Seatwork: Solve the following

 

            1. A tank initially contains 50 gal. of fresh water. Brine (water in which a certain number of pounds of salt have been dissolved), containing 2 lb/gal of salt, flows into the tank at the rate of 2 gal/min and the mixture kept uniform by stirring runs out at the same rate. How long will it take to increase from 40 lbs. to 80 lbs.?                                                                                                    ans: t = 27’ 28”

 

Homework: Solve the following

 

            1. A tank  contains 100 liters of fresh water and brine containing 2 kg of salt per liter, flows into the tank at the rate of 3 liters per min and the mixture, kept uniform by stirring, flows out at the same rate.  How many kg of salt are there in the tank at the end of 30 minutes.                            ans.  118.7kg

 

            2. Into a 100 - gal tank initially filled with brine containing 50 lb of salt flow 3 gal/min of brine containing 2 lb/gal of salt and the solution, kept uniform by stirring, flows out at the rate of 2 gal/min. How much salt is in the tank at the end of 100 minutes?                                                                                    ans.  350 lb

 

            3. A sewage disposal plant has a big holding tank of 100,000-gal  capacity. It is  Ύ  filled with liquid to start with and contains 60,000 lb of organic material in suspension. Fresh water runs into the tank at the rate of 20,000 gal/hr and the well-stirred mixture leaves at the rate of 15,000 gal/hr.  How much organic material is in the tank at the end of 3 hours?                ans.  34,722.2 lbs

 

            4. A tank initially holds 80 gal of brine containing 1/8 lb of salt per gallon. Another brine solution, containing 1 lb/gal of salt, is poured into the tank at the rate of 4 gal/min and the well-stirred mixture leaves at the rate of 8 gal/min. Find the amount of salt in the tank  a) at any time,  b) after 3 minutes and c) the time when the tank will hold 40 gallons of solution.

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 19

Simple Electric Circuits

Elementary Applications of First–Order, First Degree

Ordinary Differential Equations

 

 

6. Simple Electric Circuits

 

            Only simple electric circuits containing a resistor and an inductor or a capacitor in series with an electromotive force is considered in this section. Kirchhoff’s Voltage Law will be used to solve the problems.

 

            Kirchhoff’s Voltage Law states that the algebraic sum of the voltage drops in a simple closed electric circuit is zero.

 

 

Figure:                     R

Text Box: I

Text Box: E

 L

 

 

 

 

            Let       t    (seconds)    =   time

                        Q (coulombs)   =   quantity of electricity; e.g. charge on a capacitor

                        i    (amperes)   =   current,  time rate of flow of electricity

                        E         (volts)    =   electromotive force or voltage

                        R         (ohms)  =    resistance

                        L        (henrys) =    inductance

                        C       (farads)  =    capacitance

 

Thus, 

 

            1. If the circuit contains resistance and inductance only (R-L circuit), the

                             differential equation will be

 

                                                L  +  Ri  = E

 

            2.  If the circuit contains resistance and capacitance (R–C circuit), the

      differential equation will be

 

                        R  +  Q = E

      

Example: Solve the following:

 

            1. A resistance and inductance are connected in series in a circuit containing an impressed voltage of 100 V. If R = 10 ohms, L = 2 henries and i = 0 when t = 0, find i when t = .02second.                         

                                                                                                ans.  0.95 amp

 

            2. A resistance of 3 ohms and an inductance of 2 henries are connected in series with an electromotive force of 8e-.0001t volts. When will the current be 0.8 amperes if no current flows initially?                               

                                                                                                ans.   0.237 sec

 

 

Seatwork: Solve the following

 

            1. An inductance of 2 henries and a variable resistance r =  ohms are connected in series with a constant emf of E volts. Ifi = 0 when  t =0 and    i = 50 amp when t = 5 seconds, find the emf ,  E .                

                                                                                                ans.  28.42 V

 

 

Homework: Solve the following

 

            1. When a simple electric circuit, containing no condensers but having inductance and resistance, has the electromotive force removed, the rate of decrease of the current is proportional to the current. The current is i amperes t seconds after the cut-off,and i = 40 amperes when    t = 0. If the current dies down to 15 amperes in 0.01 sec, find i in terms of  time, t .

                                                                                                ans.  I =

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 20

Newton’s Second Law of Motion

Elementary Applications of First–Order, First Degree

Ordinary Differential Equations

 

 

7. Newton’s Second Law of Motion

 

            The unbalanced force acting on the body is proportional to the product  of the mass and its acceleration and is in the direction of the acceleration.

 

P

 

f

 

                                                            w

 

            Let F be the unbalanced force

 

                        F a ma or  F = kma where k = 1 based on experiment

           

                                    but F = P – f , m = w/g and a = dv/dt

 

                        hence

 

                                    P – f = w/g ( dv/dt)

 

                                                               Where:            F  =  force

                                                                                       m  =  mass

                                                                                       v  =  velocity

                                                                                       a =  acceleration

                                                                                        t   =  time

                                                                                       g  =  acceleration due to gravity

                                                                                                32 ft/ sec2 or 9.8

 

Example: Solve the Following

 

  1. A wagon weighing 256 lbs is pulled by a constant force of 80 lbs along a smooth road. Friction between the wagon and the smooth road is negligible but there is a wind resistance equivalent to twice the instantaneous velocity of the wagon expressed in lbs. If the wagon starts from rest, find:
    1. Velocity of the wagon after 4 sec
    2. The distance traveled by the wagon after 4 sec.
    3. The limiting velocity (t = ∞ )

 

            2. A body weighing 1960 N is pulled by a constant force of 492 N along a horizontal plane where the coefficient of friction between the body and the plane is 0.20. There is a wind resistance equal to twice the instantaneous velocity. Determine the velocity after 20 seconds.                                   ans.  9.06 m/sec

 

 

 

Seatwork: Solve the following

 

            1. A body falls from rest against a resistance that varies directly as the velocity. If the limiting speed or terminal velocity is 160 ft/sec, find the speed after  5 seconds.  Assume g = 32.                                                                                                                                                                                                ans.  101. 14 ft/sec

 

 

Homework: Solve the following

 

            1. A weight, W lb, slides down an inclined plane that makes an angle  with the horizontal. Assume that no force other than gravity is acting on the body, that is, there is no friction, no air resistance etc. At time t = 0, let the distance traveled x be x0 and let the initial velocity be v0. Determine x for t0.

                                                                                                ans. x =

 

 

 

 

 

           

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 21

Higher–Order, First Degree Differential Equations

(Linear Differential Equations with Constant Coefficients)

 

 

Standard Form:

 

                       

 

                        where a0, a1, a2. . . an–1 are constants and R(x) is a function of x.

 

 

Preliminary Theory

 

1. Initial–Value and Boundary Value Problems

 

For a linear differential equation, an nth–order initial–value problem is

 

                        Solve:

                        Subject to:

 

            Another type of problem in which the dependent variable y or its derivatives are specified at different points is called a boundary–value problem.

 

                        Solve:  

                        Subject to:  

 

The prescribed values and are called boundary conditions.

 

2. Homogeneous Equations

 

Standard Form:

    

                        ––– (1)

 

                        where a0, a1, a2, . . ., an – 1  are constants and R(x) = 0

 

            (1) is a homogeneous linear differential equation since the degree of the derivatives of each term is of first degree.

 

3. Non–Homogeneous Equations

 

Standard Form:

 

                       

 

where a0, a1, a2. . . an–1 are constants and R(x) is a function of x.

 

 

4. Differential Operators, D–Operators

 

 D–Operator

 

            An operator is a symbol indicating an operation to be performed. For D–operator, it means taking the derivative of a function with respect to x. Thus,

 

                                   

 

Algebraic Properties of D–Operator

 

1. Commutative with respect to Addition/Multiplication

                       

2. Associative with respect to Addition/ Multiplication

                       

 

3. Distributive with respect to Addition

                       

 

 

4.

        where Q(D) is a polynomial in D

 

 

Example: Perform the indicated operations:

 

            1. (D – 2) x3                                                     ans: x2 (3 – 2x) 

 

            2. (D + 1)2 xe –x                                             ans: 0

 

            3. (D2 – 36) sin 2x                             ans: –40sin 2x

           

            4.  D8 (D – m) emx                                       ans: 0

 

Seatwork: Perform the indicated operations:

 

            1. D2(2x3)                                           ans: 12x

 

            2. (D2 – 4D + 3)(3x2 + 2 cosx)         ans: 9x2 – 24x + 6 + 8sinx + 4 cos x

 

           

Homework: Perform the indicated operations:

 

1.      D(x lnx)                                         ans: 1 + lnx

 

2.      D(x2 + sin 4x – lnx)                      ans: 2x + 4 cos 4x – 1/x

 

3.      (2D2 – 3D + 5)(x cosx – 3)         ans: 3xsinx + 3xcosx –4sinx –3cosx - 15

 

            4. (D3 + 2D – 4) (e-xsinx + e2x)         ans: 4e-x(cosx - sinx) + 8e2x

           

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 22

Homogeneous Linear Equations with Constant Coefficient

 

 

            To find its general solution, y = yc + yp where the complementary solution, yc, may be determined from the roots of the auxiliary solution, f(m) = 0 and the particular solution, yp, is zero since R(x) = 0. Therefore y = yc

 

 

                        Recall:

                                   

 

                        Using D–operators,

 

                                   

 

Standard Form:

 

                                    ––––––– (1)

 

 

Suppose y = emx is a particular solution of (1). Substitute it in (1),

 

                                   

 

                                   

 

 

                        Recall:

 

                                   

 

                                               

 

                                                                       

 

            For emx to be a particular solution of (1), it is necessary and sufficient that  f (m) = 0. This relation is called the characteristic or auxiliary equation of (1). The roots of the auxiliary equation, f(m) = 0. Determine the general solution of (1) as follows:

 

1. The Auxiliary Equation with Real Roots

 

1.1 Distinct Roots

 

            Suppose the auxiliary equation, f(m) = 0, has m1, m2, and m3 as its roots,  then em1x, em2x  and em3 x are the particular solution of (1) and the general solution is given by:

 

                                               

 

Example: Find the general solution of the differential equation.

           

            1. (D2 – 2D – 3) y = 0           

           

            2. (D2 + 5D + 4) y = 0           

           

Seatwork: Find the general solution of the differential equation

           

            1. (D3 – 7D + 6) y = 0            ans:  y  =  C1e+ C2e2x  +  C3e-3x

 

            2. (D5 – 9D4 + 13D3 + 57D2 – 86D – 120) y = 0              

                                                ans:  y  =  C1e-x  +  C2e-2x  +  C3e3x   +  C4e4x  +  C5e5x                                        

                                                           

1.2 Repeated Roots

 

            Suppose the roots of the auxiliary equation, f(m) = 0, are m1 (taken twice as a root) and m2 (taken 4 times as a root) so that each term could be linearly independent, then the general solution is given by:

 

 

                       

 

Example: Find the general solution of the differential equation.

 

            1. (D2 + 4D + 4) y = 0                                               

 

            2. (D3 + 3D2 + 3D + 1) y = 0                        

 

Seatwork: Find the general solution of the differential equation

 

            1. (D3 – 4D2 + 4D) y = 0       ans:  y  =  C1 +  (C2 +  C3x) e2x                    

           

            2. (D4 – 6D3 – 24D2 + 224D – 384) y = 0             

                                                            ans:  y  =  C1e-6x  +  (C2 +  C3x  + C4x2)  e4x

2. The Auxiliary Equation with Complex/Imaginary Roots

 

2.1 Distinct Roots           

 

            Suppose the roots of the auxiliary equation, f(m) = 0, are m1 = a + bi and m2 = a – bi, then the general solution is given by:

 

                                               

Example: Find the general solution of the differential equation.

 

            1. (D2 + 6D + 13) y = 0        

 

            2. (D2 + 4D + 5) y = 0           

 

Seatwork: Find the general solution of the differential equation

 

1.      (D4 + 10D2 + 16) y = 0

                        ans:  y=(C1cos 2Φ2 x + C2sin 2Φ2 x)+(C3cosΦ2 x +C4sin Φ2 x)

 

2.2 Repeated Roots

 

            Suppose the roots of the auxiliary equation, f(m) = 0, are m1 = a + bi (taken twice as a root) and m2 = a – bi (taken twice as a root), then the general solution is given by:

 

                                   

Example: Find the general solution of the differential equation.

 

            1. (D2 – 2D + 5)2 y = 0                     

 

            2. (D4 + 18D2 + 81) y = 0                 

 

Seatwork: Find the general solution of the differential equation

 

            1. (D4 – 4D3 + 8D2 – 8D + 4) y = 0 

                        ans:  y  = [(C1 +  C2x) cos x  +  (C3 +  C4x) sinx] ex

           

            2. (D6 + 3D4 + 3D2 + 1) y = 0          

                        ans:  y  =  [(C1 +  C2x  +  C3x2) cos x  +  (C4 +  C5x  +  C6x2) sin x]

 

 

 

 

 

 

 

Lesson 23

The Method of Undetermined Coefficients

Non–Homogeneous Equations

 

 

Methods to be used in solving non-homogeneous equations:

1.      The Method of Undetermined Coefficients

2.      Variation of Parameters

3.      Inverse Operators

4.      By Inspection

 

 

Standard Form:

 

                       

 

            To find its general solution, y = yc + yp. The complementary solution, yc, may be determined from the roots of the auxiliary equation, f(m) = 0, and the particular solution, yp, may be determined by:

 

 

1. The Method of Undetermined Coefficients

 

 

Step 1.            Consider the right hand side of the equation, R(x) and determine its                 roots.

 

            Case 1.          If there is no repetition of roots between the auxiliary                                                       equation and R(x) such that

           

 

 

when R(x) is a                                        then yP is      

           

            a. constant like 2, 3, etc.                                           A

           

            b. linear function like 3x, 2x + 5, etc.                       Ax + B

           

            c. quadratic function like x2, x2 – 3x, etc.                 Ax2 + Bx +  C

           

            d. xn                                                                                                      Axn + Bxn – 1 +...+ Dx + F

           

            e. eax                                                                                                    Aeax

           

            f. sin ax, cos ax, sin ax  +  cos ax                            A sin ax + B cos ax

 

           

 

            Case 2.          If there is a repetition of roots between the auxiliary equation                             and R(x), follow the principles discussed in the                                                                     homogeneous linear differential equation under repetition of                                                 roots.

 

Step 2.            Differentiate yP according to the order of the given equation.

 

Step 3.            Substitute these equations in the given equation. Solve for the                          literal constants by comparing/collecting their coefficients.

           

 

Example: Find the general solution of the differential equation.

           

            1. (D2 – D – 2) y = 5                                     

 

            2. (D2 + 4) y = 4x + 3                        

 

            3. (D2 + 2D + 2) y = 2x2 + 3x + 8

 

            4. (D2 + 4D + 4) y = 2e2x                              

 

            5. (D2 + 9) y = 10 sin 2x + 5 cos 2x

 

 

Seatwork: Find the general solution of the differential equation.

 

            1. (D2 – 3D + 2) y = 2x3 – 4x2 + 6x + 3                                                                                           ans.

 

            2. (D2 – 9) y = x + e2x – sin 2x                                                                                                          ans.  

 

Homework: Find the general solution of the differential equation.

 

            1. (D2 – 2D – 3) y = x3 + sin x

                        ans.

           

            2. (D2 – 2D) y = ex sin x                                

                        ans:

           

            3. (D2 – 4D + 4) y = x3e2x + 2xe2x                                                                                                    ans.

           

            4. (D2 – 9) y = 2x + 3e2x – sin x

 

            5. (D2 + 1) y = –2 sin x + 4x cos x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 24

Variation of Parameters

Non-Homogeneous Equations

 

 

Step 1.            Change the constants in the yC to a function of the independent                                     variable (parameters), say A, B, etc. where A, B are functions of x.

 

Step 2.            Differentiate yP according to the order of the given equation setting                              the sum of all functions of the derivatives of the independent                                                 variable to zero and the last nth derivative of the independent                                  variable to R(x).

 

Step 3.            By solving simultaneously these equations, find the parameters.

 

 

Example: Find the general solution of the differential equation.

 

            1. (D2 – D – 2) y = 5

 

            2. (D2 – 1) y = ex + 1            

 

            3. (D3 + D) y = csc x 

                                    ans:  y  =  C1 +  C2cos x  +  C3sinx  +  ln (csc x - cot x)  -                                                                          cosx ln (sin x)  -  x sinx

 

 

Seatwork: Find the general solution of the differential equation.

            1. (D2 – 6D + 9) y =                   

                        ans: y = C1 + C2 cos x + C3 sin x + ln (csc x – cot x)

                                                – cos x ln (sin x) – x sin x

 

Homework: Find the general solution of the differential equation.

 

            1. (D2 -4)y = 4x – 3ex            ans: y = ex – x + c1e2x + c2e-2x

 

            2. (D3 – 2D2 + d)y = x            ans: x2/2 + 2x + 3 + c1xex + c2ex + c3

 

 

 

 

 

 

 

 

Lesson 25

R(x) is an Exponential Function

Inverse Operator

 

 

            Consider the equation,

                                                F(D) yP = R(x)  ––––––– (1)

 

            Applying the inverse operator  in  (1),

                                                            yP = R(x)

 

Note:  yP depends on the type of function represented by R(x).

 

R(x) may be:

1.      Exponential Function

2.      Trigonometric Function

3.      Polynomial Function

4.      Composite Function

4.1   Exponential Shift

4.2   X Shift

 

 

 Properties of Inverse Operators:

 

            Let and be inverse operators.

           

                        1. =

           

                        2. =

           

                        3.

 

 

 

 

 

 

 

1.  R(x) is an Exponential Function

 

            1. Given: F(D) yP = eax

           

                            Such that F(a) Ή 0

 

            2. Given: (D – a)m y = eax

           

                          Such that F(a) = 0

 

 

Example: Find the particular solution of the differential equation.

 

            1. (D2 – D – 2) yp = 5                        ans: yp = -5/2

 

            2. (D2 + 4D + 4) yp = 2e2x                 ans: yp = 1/8 (e2x)

 

 

Seatwork: Find the particular solution of the differential equation.

 

            1. (D3 – 2D2 – 5D + 6) yp = (e2x + 3)2         

                                                            ans: yp = 1/18 (e4x) – 3/2 (e2x) + 3/2           

 

 

            2. (D3 – 5D2 + 8D – 4) yp = e2x + 2ex + 3e–x          

                                                            ans: yp = x2/2 (e2x) + 2xex – 1/6 (e-x)

 

 

Homework: Find the particular solution of the differential equation.

 

1.      (D2 – 2D – 3) yp = e-2x + ex + 6

 

2.      (D2 + 2D + 1) yp = 2 sinh 2x

 

3.      (D2 – D – 2) yp = e2x + e-x

 

 

 

 

 

 

 

Lesson 26

R(x) is a Trigonometric Function

Inverse Operator

 

 

2.  R(x) is a Trigonometric Function

           

            2.1 Given:  F(D2) yP = sin ax or F(D2) yP = cos ax

 

a.     

b.     

 

                                         Such that F(–a2) Ή 0

 

            2.2  Given: (D2 + a2) y = sin ax or (D2 + a2) y = cos ax

 

a.     

 

b.               Such that F(–a2) = 0

 

 

Example: Find the particular solution of the differential equation.

 

            1. (D2 + 9) yp = 10 sin2x + 5 cos 2x ans: yp = 2sin2x + 5 cos2x

 

            2. (D2 + 3D – 4) yp = sin 2x                          ans: yp = -1/50(3cos2x + 4sin2x)

 

            3. (D2 + 4) yp = cos 2x + cos 4x                   ans: yp =Ό(x sin2x) – 1/12(cos4x)

 

 

Seatwork: Find the particular solution of the differential equation.

 

            1. (D2 + 4) yp = sin 3x                                    ans:  yP =

            2. (D4 + 10D2 + 9) yp = cos (2x + 3)            ans:  yP =

 

 

 

 

Homework: Find the particular solution of the differential equation.

 

1.      (D2 + 16) yp = sin 2x + cos 2x                ans: yp = 1/12(sin2x) + 1/7(cos3x)

 

2.      (D2 + 9) yp = 2 sin2x                                 ans: yp = 1/9 – 1/5(cos2x)

 

3.      (D – 6) yp = sin 2x                        ans: yp = -1/20(cos2x) -3/20(sin2x)

 

4.      (D2 + D – 2) yp = sin 2x                           ans: yp = -1/20(cos2x + 3sin2x)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 27

R(x) is a Polynomial Function

Inverse Operator

 

 

3.  R(x) is a Polynomial Function

 

            Given:    F(D) yP = xn

 

 

Note: is obtained by expanding in ascending powers of D and suppressing all terms beyond Dm since Dm (xn) = 0 when m > n.

 

 

Example: Find the particular solution of the differential equation.

 

1.      (D2 + 4) yp = 4x + 3                                  ans: yp = x + 3/4

 

2.  (D2 + 2D + 2) yp = 2x2 + 3x + 8               ans: yp = x2 – ½(x) + 3/2

 

 

Seatwork: Find the particular solution of the differential equation.

 

            1. (2D2 + 2D + 3) yp = x2 + 2x – 1                ans: 

            2. (D3 – 4D2 + 3D) yp = x2                             ans:

 

Homework: Find the particular solution of the differential equation.

 

1.      (D2 + 1) yp = x6                                         ans: yp = x6 - 30x4 + 360 x2 - 720

 

2.      (D2 + D – 2) yp = x2                                  ans: yp = - ½ ( x2 + x + 3/2)

 

 

 

 

 

 

 

 

Lesson 28

R(x) is a Composite Function

Inverse Operator

 

 

4.  R(x) is a Composite Function

 

            4.1 Given:

 

 

 

                        Using “exponential shift”

 

 

            4.2 Given:

 

                        Using “x–shift”

 

 

 

Example: Find the particular solution of the differential equation.

 

            1. (D2 - 2D) yp = ex sin x                    ans: yp = -1/2 (exsinx)

 

            2. (D2 + 2D + 4) yp = ex sin 2x          ans: yp = -1/73(ex)(8cos2x - 3sin2x)

 

            3. (D2 + 3D + 2) yp = x sin 2x            ans: yp1 = -1/20(x) (3cos2x + sin2x)

                                                                                yp2 = -1/200(24sin2x + 7cos2x)

 

 

Seatwork: Find the particular solution of the differential equation.

 

            1. (D2 – 4) yp = x2e3x                         

                        ans:

           

            2. (D3 – 3D2 – 6D + 8) y = xe–3x      

                        ans:

 

 

Homework: Find the particular solution of the differential equation.

 

1.      (D – 2)2 yp = x e2x                                    ans: yp = 1/6 (x3e2x)

 

2.      (D2 – 4D + 13) yp = e2x sinx                    ans: yp = 1/8 (e2x sinx)

 

3.      (D2 + 9) yp = x sin 2x                               ans: yp = 1/25(5xsin2x – 4cos2x)

 

4.      (D + 1) yp = x cosx                       ans: yp = x/2(sinx + cosx) – ½(sinx)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 29

By Inspection

 

            It is easy to obtain a particular solution of a non–homogeneous linear differential equation by inspection if R(x) is a constant, R0.

 

            Standard Form:

                       

                                   

           

                        Case I.

 

                        Case II.

           

            Where:            R0 = constant found on the right side of the equation

                        an = constant term

                        an–k = coefficient of the lowest–ordered derivative

                        k = lowest–ordered derivative

 

Example: Find the particular solution of the differential equation.

 

            1. (D2 + 4) yp = 12

           

            2. (D2 – D – 2) yp = 5           

 

            3. (D5 – 9D3) yp = 81

 

Seatwork: Find the particular solution of the differential equation.

 

1.      (D2 – 2D + 8) yp = 16

 

2.      (D3 + D2 – 2D) yp = 20

 

Homework: Find the particular solution of the differential equation.

 

1.      (D2 – 9D - 3) yp = 27

 

2.      (D4 – 2D3 - 5D2 + 6D) yp = 15

 

 

 

Lesson 30

The Laplace Transform

 

 

            Let f(t) be a function of t defined for each positive values of t. Then the Laplace transform of f(t), donated by , is defined by

                                     

                                    ––––––– (1)

 

provided that the integral exists. s is a parameter, which may be a real or complex number.  being clearly a function of s is briefly written as F(s). That is,

 

                                   

 

                                    where or – Laplace Transform of f(t).

 

Transforms of Elementary Functions

           

            The Laplace transforms of some exponential, trigonometric, polynomials and hyperbolic functions are:

           

            1.

                       

 

            2.

                       

 

            3.

 

            4.

                       

 

Properties of Laplace Transforms

 

1.      Linearity Property

 

            If a, b, c be any constants and f, g, h be any functions of t, then

 

                       

 

2.      First Shifting Property

 

                        If , then

 

                                   

 

Application of this property leads us to the following useful results:

 

1.     

 

2.     

 

3.     

 

4.     

 

5.     

 

                        where in each case, s > a.

 

Example: Find the Laplace transform of the following:

 

            1. f(t) = 2e2t – 3 sin 3t + 7 cos 3t + 5t3

           

            2. f(t) = t3 e–3t

           

            3. f(t) = e–2t sin 4t

           

Seatwork: Find the Laplace transform of the following:

            1. f(t) = e–3t(2 cos 5t – 3 sin 5t)                    ans:

 

            2. f(t) = sin 2t sin 3t                                       ans:

 

Homework: Find the Laplace transform of the following:

 

1.      f(t) = te–4t sin 3t

 

2.      f(t) = 2t2 + sint + 2e3t

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Lesson 31

Inverse Laplace Transforms

 

 

            If represents the Laplace transform of a function, that is, then is the inverse Laplace transform of , that is, .

 

 

Table of Inverse Laplace Transform

            1.

            2.

            3.

            4.

            5.

            6.

            7.

            8.

            9.  

            10.

            11.

            12.

 

            When evaluating inverse transforms, it often happens that a function of s under consideration does not match exactly the form of a Laplace transform  given in a table. It may be necessary to “fix up” the function of s by multiplying and dividing by an appropriate constant.

 

Linearity Property:  is a Linear Transform, that is, for constants:

                       

 

            Partial fractions play an important role in finding Laplace transforms when the denominator of f(s) is factorable into distinct linear factors.

 

 

Example: Evaluate the following

 

1.                                             ans:

2.                                    ans:

3.                                 ans:  

4.                ans:

 

 

Seatwork: Evaluate the following:

 

           1.

 

 

Homework: Evaluate the following

 

            1.

 

            2.

 

 

 

 

References:

 

Asin, Ricardo C. , Differential Equations, A Handy Reference Book for College, Tru-Copy Publishing House, Inc. 1997

 

Bautista, Lincoln et.al. Differential Equations. Philippines, 1999.

 

De La Fuente, Ruben A. et al, Elementary Differential Equations, Reference Text, Merriam and Webster Bookstore, Inc, Manila, Philippines, 1999

 

Mateo, Rolando A. et al. Differential Equations, MRTTC

 

Rainville, Earl D. and Bedient, Phillip E. Elementary Differential Equations. Macmillan Publishing Company. 8th ed.

 

Richard Bronson, Ph.D., Schaum’s Solved Problems Series, Differential Equations, Mc Graw Hill Book Company, USA

 

Zill, Dennis G; A First Course in Differential Equations with Modeling Applications; Brooks/Cole. 7th ed., 2000.