The problems that I had solved are contained in "Introduction to ordinary differential equations (4th ed.)" by Shepley L. Ross

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PROBLEM SET & SOLUTIONS

DIFFERENTIAL EQUATION

By: Ibnu Rafi

e-mail: ibnu257fmipa@student.uny.ac.id

Page 2 of 72

Table of Contents

Table of Contents ................................................................................................................................................... 2

Solution of Exercise 1 (Linear and Nonlinear D.E)................................................................................ 3

Solution of Exercise 2 (Ordinary and Partial D.E) ................................................................................. 4

Solution of Exercise 3 (Solution of D.E) ...................................................................................................... 6

Solution of Exercise 4 (Initial Value Problem) ..................................................................................... 11

Solution of Exercise 5 (Separable D.E) ..................................................................................................... 13

Solution of Exercise 6 (General Solution of Separable D.E) ........................................................... 15

Solution of Quiz .................................................................................................................................................... 21

Solution of Exercise 7 (Homogeneous D.E) ........................................................................................... 22

Solution of Exercise 8 (Solution of Homogeneous D.E) ................................................................... 23

Solution of Exercise 9 (Non Homogeneous D.E) ................................................................................. 28

Solution of Exercise 10 (Solution of Non Homogeneous D.E) ...................................................... 29

Solution of Exercise 11 (Exact and Non Exact D.E) ............................................................................ 35

Solution of Exercise 12 (Integrating Factor) ......................................................................................... 42

Solution of Exercise 13 (Grouping Method) .......................................................................................... 44

Solution of Quiz .................................................................................................................................................... 45

Solution of Exercise 14 (Linear D.E) ......................................................................................................... 48

Solution of Exercise 15 (Solution of Linear D.E) ................................................................................. 50

Solution of Exercise 16 (Properties of Linear D.E) ............................................................................ 51

Solution of Exercise 17 (Integrating Factor of Linear D.E) ............................................................ 53

Solution of Exercise 18 (Orthogonal and Oblique Trajectories).................................................. 57

Solution of Exercise 19 (Problem in Mechanics (Frictional Forces)) ....................................... 64

Solution of Exercise 20 (Rate Problems (Rate of Growth and Decay and Population)

Growth) ................................................................................................................................................................... 66

Solution of Exercise 21 (Mixture Problem) ........................................................................................... 68

Solution of Exercise 22 (Reduction of Order) ....................................................................................... 70

Page 3 of 72

Solution of Exercise 1 (Linear and Nonlinear D.E)

We will determine whether the equations under consideration is linear or nonlinear. To

determine whether the equations under consideration is linear or nonlinear we should

know that differential equation are said to be nonlinear if any product exist between the

dependent variable and its derivatives, between the derivatives themselves, or the

dependent variable is trancedental function.

1. 2

 + 5 

 + 6 = 0 (Linear differential equation)

Since we see that the dependent variable of the differential equation above is

and its various derivatives occur to the first degree only.

2. 4

4+ 2 3

3+ 3 

 =  ( Linear differential equation)

Since we see that the dependent variable of the differential equation above is

and its various derivatives occur to the first degree only.

3. 2

2 + 5 

 + 6 2 = 0(Nonlinear differential equation)

Since we see that the dependent variable of the differential equation above is

and its various derivatives occur to the first degree only, but then, the differential

equation above contains the product between the dependent variable

themselves, that is in the term 6 2. Therefore, the differential equation

2

2 + 5 

 + 6 2 = 0 is nonlinear differential equation.

4. 2

2 + 5 

 3

+ 6 = 0 (Nonlinear differential equation)

Since we see that the dependent variable of the differential equation above is ,

but in the term 5 

 3

involves the third degree of the first derivative or in the

othere word there is product between three derivatives. Therefore, the

differential equation 2

2 + 5 

 3

+ 6 = 0 is nonlinear differential equation.

5. 2

2 + 5 

 + 6 = 0 (Nonlinear differential equation)

Since we see that the dependent variable of the differential equation above is ,

but there is product between the dependent variable and its derivative in the

term 5

 . Therefore, the differential equation 2

2 + 5 

 3

+ 6 = 0 is

nonlinear differential equation.

Page 4 of 72

Solution of Exercise 2 (Ordinary and Partial D.E)

We will classify of the following differential equations as ordinary or partial differential

equations, state the order of each equation, and determine whether the equation under

consideration is linear or nonlinear.

Ordinary differential equation is the differential equation involving ordinary

derivatives of one or more dependent variables with respect to a single

independent variable.

Partial differential equation is the differential equation involving ordinary

derivatives of one or more dependent variables with respect to more than one

independent variable.

The order of a differential equation is equal to the order of the highest

differential coefficient that it contains.

The degree of a differential equation is the highest power of the highest order

differential coefficient that the equation contains after it has been rationalized.

1. 

 + 2 = .

The differential equation 

 + 2 = is ordinary differential equation (since

it has only one independent variable, that is ), first order ordinary differential

equation, 

 , first degree ordinary differential equation, and linear differential

equation (since no product between dependent variable ( ) themselves, no

product between and/or any of its derivatives, and its various derivatives

occur to the first degree only).

2. 3

3 + 4 2

2 5 

 + 3 = sin .

The differential equation 3

3 + 4 2

2 5 

 + 3 = sin is ordinary differential

equation (since it has only one independent variable, that is ), third order

ordinary differential equation, 3

3 , first degree ordinary differential equation,

and linear differential equation (since no product between dependent variable

( ) themselves, no product between and/or any of its derivatives, its various

derivatives occur to the first degree only, and no trancendental function of

and/ or its derivatives occur).

3. 2

2+ 2

2 = 0.

The differential equation 2

2+ 2

 2 = 0 is partial differential equation (since it

has more that one independent variable involved, that is and ), second order

Page 5 of 72

partial differential equation, first degree partial differential equation, linear

differential equation (since no product between dependent variable ( )

themselves, no product between an d/or any of its derivatives, and no

trancendental function of and/ or its derivatives occur).

4. 2  + 2  = 0.

The differential equation is 2  + 2  = 0 

 = 2

2 ordinary differential

equation (since it has only one independent variable involved, that is either or

), first order ordinary differential equation, first degree ordinary differential

equation, nonlinear differential equation (since there is a product either between

dependent variable themselves ( if is dependent variable) or between

dependent variable themselves ( if is dependent variable) ).

5. 4

4 + 3 2

2 5

+ 5 = 0.

The differential equation 4

4 + 3 2

2 5

+ 5 = 0 is ordinary differential

equation (since it has only one independent variable involved, that is ), fourth

order ordinary differential equation, first degree ordinary differential equation,

nonlinear differential equation (since in term 3 2

2 5

involves the fifth power of

the second derivatives ).

6. 4

2 2+ 2

2+ 2

2+= 0.

The differential equation 4

2 2+ 2

2+

2+= 0 is partial differential

equation (since it has more than one independent variable involved, that is and

), fourth order partial differential equation ( 4

2 2 is the fourth derivatives of

, ), first degree partial differential equation, linear differential equation

(since since no product between dependent variable () themselves, no product

between and/or any of its derivatives, and no trancendental function of and/

or its derivatives occur).

7. 2

2+sin = 0.

The differential equation 2

2+sin = 0 is ordinary differential equation

(since it has only one independent variable, that is ), second order ordinary

differential equation, 2

2 , first degree ordinary differential equation, and linear

differential equation (since no product between dependent variable ( )

themselves, no product between and/or any of its derivatives, and no

trancendental function of and/ or its derivatives occur).

Page 6 of 72

8. 2

2+sin = 0.

The differential equation 2

2+sin = 0 is ordinary differential equation

(since it has only one independent variable, that is ), second order ordinary

differential equation, 2

2 , first degree ordinary differential equation, and

nonlinear differential equation (since its dependent variable is , but then, there

is trancendental function of occur in term sin ).

9. 6

6+ 4

 4 3

 3 + = .

The differential equation 6

6+ 4

 4 3

 3 + = is ordinary differential

equation (since it has only one independent variable, that is ), sixth order

ordinary differential equation, 6

6 , first degree ordinary differential equation,

and nonlinear differential equation (since there is a product between the various

derivatives of with respect to in the term 4

 4 3

3 ).

10. 

 3

= 2

2 + 1

The differential equation 

 3

= 2

2 + 1 is ordinary differential equation

(since it has only one independent variable, that is ), second order ordinary

differential equation, , first degree ordinary differential equation, and nonlinear

differential equation (since in term 

 3

involves the third power of the first

derivatives).

Solution of Exercise 3 (Solution of D.E)

1. a) We will show that = + 3 is a solution of the differential equation that be

defined as 

 += + 1 on every interval < < of the axis. To show this, we

must show that = + 3 satisfies the differential equation 

 += + 1

( ordinary linear nonhomogen differential equation). By differentiating , we

obtain = 1 3 ,  , . Afterwards, subtituting ( ) for , for 



in the mentioned differential equation. We obtain



 += 1 3  + + 3  = 1 3  ++ 3  =+ 1,

that is, + 1 = + 1. Therefore, the given differential equation is satisfied by

= + 3  . In the other word, = + 3  is a solution of the differential

equation that be defined as 

 += + 1 on every interval < < of the 

Page 7 of 72

b) We will show that = 2 3 5 4 is a solution of the differential equation

that be defined as 2

2 7 

 +12 = 0 on every interval << of the axis. To

show this, we must show that = 2 3 5 4 ,  , satisfies the

differential equation 2

2 7 

 +12 = 0. By differentiating ( ), we obtain

= 6 3 20 4 and  =18 3 80 4 . Afterwards, subtituting ( ) for

, ( ) for 

 , and  () for 2

2 in the mentioned differential equation. We obtain

2

 2 7 

 +12 = 0

18 3 80 4 7 6 3 20 4 +12 2 3 5 4 = 0

18 3 80 4 42 3 +140 4 +24 3 60 4 = 0

18 42 +24 3 + 80 + 140 60 4 = 0

0. 3 + 0. 4 = 0

0 3 +4 = 0

0 = 0.

Therefore, the given differential equation is satisfied by = 2 3 5 4 . In the

other word, = 2 3 5 4 is a solution of the differential equation that be

defined as 2

2 7 

 +12 = 0 on every interval << of the 

c) We will show that = + 2 2+ 6 + 7 is a solution of the differential

equation that be defined as 2

2 3 

 + 2 = 4 2 on every interval < < of the

axis. To show this, we must show that = + 2 2 + 6 + 7,  ,

satisfies the differential equation 2

2 3 

 + 2 = 4 2 . By differentiating ( ), we

obtain = + 4 + 6 and  = + 4. Afterwards, subtituting ( ) for ,

( ) for 

 , and  () for 2

2 in the mentioned differential equation. We obtain

2

 2 3 

 + 2 = 4 2

+ 4 3 + 4 + 6 + 2 + 2 2 + 6 + 7 = 4 2

+ 4 3 12 18 + 2 + 4 2 +12 +14 = 4 2

1 3 + 2 + 4 2 + 12 +12 + 418 +14 = 4 2

0. + 4 2 + 0. + 0 = 4 2

0 + 4 2 + 0 + 0 = 42

Page 8 of 72

4 2 = 4 2 .

Therefore, the given differential equation is satisfied by = + 2 2 + 6 + 7. In

the other word, = + 2 2+ 6 + 7 is a solution of the differential equation

that be defined as 2

2 3 

 + 2 = 4 2 on every interval < < of the axis

d) We will show that = 1

1+ 2 is a solution of the differential equation that be

defined as 1 + 2 2

2 + 4 

 + 2 = 0 on every interval < < of the axis. To

show this, we must show that = 1

1+ 2 satisfies the differential equation

(1 + 2 ) + 4 + 2 = 0. By differentiating ( ), we obtain = 2

1+ 2 2 and

 = 21+ 2 2 2 2 1+ 2 2

1+ 2 4= 1+ 2  8 2 2 1+ 2 

1+ 2 4= 6 2 2

1+ 2 3. Afterwards subtitut-

ing ( ) for , ( ) for 

 , and  () for 2

2 in the mentioned differential equation.

We obtain

1 + 2  + 4  + 2 = 0

1 + 2 6 2 2

1 + 2 3 + 4  2

1 + 2 2 + 2 1

1 + 2 = 0

6 2 2

1 + 2 2 8 2

1 + 2 2 +2 + 2 2

1 + 2 2 = 0

6 8 + 2 2 + 2 + 2

1 + 2 2 = 0

0. 2 + 0

1 + 2 2 = 0

0

1 + 2 2 = 0

0 = 0.

Therefore, the given differential equation is satisfied by = 1

1+ 2 . In the other

word, = 1

1+ 2 is a solution of the differential equation that be defined as

1 + 2 2

2 + 4 

 + 2 = 0 on every interval < < of the 

2. a) We will show that 3 + 3 2 = 1 = 1 3

3 2 = 1 3

3 is an implicit solution

of the differential equation 2

 + 2 + 2 = 0 on the interval 0 < < 1. To

show this, firstly, we differentiating 3 + 3 2 = 1 implicitly with respect to . We

obtain

Page 9 of 72

3 + 3  2

 = 1



3

 + 3  2

 = 1



3 2 + 3  2

 = 0

3 2 + 3 1. 2 + 2 

 = 0

3 2 + 3 2 + 6 

 = 0



= 3( 2 + 2 )

6  (0,1) .

By subtituting (*) to 2

 + 2 + 2 = 0 we obtain

2 

+ 2 + 2 = 0

2  3( 2 + 2 )

6 + 2 + 2 = 0

2 + 2 + 2 + 2 = 0

2 +1 3

3 + 2 + 1 3

3 = 0,  (0,1)

0 = 0.

Thus, we can conclude that 3 + 3 2 = 1 is an implicit solution of the differential

equation 2

 + 2 + 2 = 0 on the interval 0 < < 1

b) We will show that 5 2 2 2 3 2 = 1 is an implicit solution of the differential

equation 

 + = 3 3 on the interval 0 < < 5

2. To show this, firstly, we

differentiating 5 2 2 2 3 2 = 1 implicitly with respect to . We obtain

5 2 2 2 3 2

 = 1



10 2 

 +10  2 4 3 

 6 2 2 = 0

10 2 4 3 

 = 6 2 2 10  2

Page 10 of 72



 =6 2 2 10  2

10 2 4 3 ,0, 5

2

and

5 2 2 2 3 2 = 1

5 2 2 3 2 = 1

=1

5 2 2 3 = 1

5 2 ,  0, 5

2 

By subtituting (*) and (**) to 

 + 3 3 = 0 we obtain

6 2 2 10 2

10 2 4 3 + 1

5 2 2 3 = 3 3

 2 3 5

2 5 2 +1

5 2 = 3 3

3 5

5 2 + 1

5 2 = 3 3

3 5

5 2 5 2 +1

5 2 = 3 3

3 5 + 5 2

5 2 5 2 = 3 3

5 2 5 2 = 3 3

1

5 2 5 2 = 3 3

3 1

3 5 2 5 2 = 3 3

3 1

5 2 3

= 3 3 , 0, 5

2

3 3 = 3 3 , 0, 5

2.

Thus, we can conclude that 5 2 2 2 3 2 = 1 is an implicit solution of the

differential equation 

 + = 3 3 

 + 3 3 = 0 on the interval

0 < < 5

2

Page 11 of 72

Solution of Exercise 4 (Initial Value Problem)

1. Show that = 4 2 + 23 is a solution of initial- value problem

2

 2 + 

 6 = 0,

0 = 6,

0 = 2.

Is = 2 2 + 43 also a solution of this problem? Explain why and why not.

Proof

Assume that = 4 2 + 23 is a solution of given initial- value problem. Since

= 4 2 + 23 , we obtain = 8 2 6 3 ,  =16 2 +18 3 , and

 + 6 = 16 2 +18 3 + 8 2 6 3 24 2 12 3 = 0 (satisfied).

Moreover, by subtituting = 0 to = 4 2 + 23 and  = 8 2 6 3 we

obtain (0) = 4 0 + 2 0 = 0 4 + 2 = 1.6 = 6 (the initial condition is satisfied)

and 0 = 8 0 6 0 = 0 8 6 = 1.2 = 2 (the second condition is satisfied).

From this result we can say that our assumption is accepted. Therefore, we can

conclude that = 4 2 + 23 is the solution of given initial- 

Afterward, we will observe whether = 2 2 + 43 is also a solution of the

given problem or not. We know that = 4 2 123 ,  = 8 2 +363

and

 +  6 = 8 2 +36 3 + 4 2 12 3 +12 2 +24 3 = 0

(satisfied). Moreover, by subtituting = 0 to = 2 2 + 43 and = 4 2

123 , we obtain

0 = 2 0 + 40 =0 2 + 4 = 1.6 = 6 (satisfied)

and 0 = 4 0 12 0 = 0 412 = 1 8 = 8 2 (unsatisfied).

From this result, we can conclude that = 2 2 + 43 is not the solution of

given problem since one of the initial conditions in initial- value problem, that is

0 = 2 not be satisfied when we subtituting = 0 to = 4 2 12 3 .

2. Every solution of the differential equation 2

2+= 0 may be written in the

form = 1 sin + 2cos , for some choice of the arbitrary constants 1 and 2.

Using this information, show that boundary problem (a) and (b) possess

solution but that (c) does not.

2

 2 + = 0, 0 = 0,

2 = 1.

Page 12 of 72

2

 2 + = 0, 0 = 1,

2 =1.

2

 2 + = 0, 0 = 0, = 1.

Solution

(a) Since every solution of differential equation 2

2+= 0 may be written in

form = 1 sin + 2cos , by subtituting = 0 and =

2 to = 1 sin +

2 cos , we obtain 0 = 1 sin 0 + 2 cos 0 = 2 = 0

and

2 = 1 sin

2 + 2 cos

2 = 1 . 1 + 2 . 0 = 1 = 1.

Therefore, the given boundary problem possess solution and it particular

solution is = sin .

(b) Since every solution of differential equation 2

2+= 0 may be written in

form = 1 sin + 2cos , by subtituting = 0 to = 1sin + 2cos we

obtain that 0 = 1 sin 0 + 2 cos 0 = 1 . 0 + 2 . 1 = 2 = 1.

Since = 1 sin + 2cos , we obtain that = 1cos 2sin . By

subtituting =

2 to = 1 cos 2 sin , we obtain

2 = 1 cos

2 2 sin

2 = 1 . 0 2 . 1 =  2 =1 2 = 1.

Since, the value of 2 consistent, that is 2= 1, we can conclude that the given

boundary problem possess solution. Moreover, the general solution for such

boundary problem is = 1 sin( ) + cos( ).

(c) Since every solution of differential equation 2

2+= 0 may be written in

form = 1 sin + 2cos , by subtituting = 0 and = to = 1sin +

2 cos , we obtain

0 = 1 sin(0) + 2 cos(0) = 1 . 0 + 2 . 1 = 2 = 0

and = 1sin + 2cos = 1. 0 + 2. 1 = 2= 1 2= 1.

Since the value of 2 inconsistent, we can say that the given boundary

problem has no solution.

Page 13 of 72

***

Solution of Exercise 5 (Separable D.E)

Determine whether each of the following differential equations is or is not separable.

Note: An equation of the form  +  = 0 ( ) is called an

equation with variables searable or simply a separable equation. Equation (*) can be

restated

 +

 = 0.

1. 

 = 3 2 2

Solution:

Since



 = 3 2 2 sin



 = 2 3 sin 

1

2  = 3 sin  

3 sin  1

2  = 0,

we can conclude that the differential equation 

 = 3 2 2 sin is separable.

2. 

 = 3 sin

Solution:

Since



 = 3 sin

 = 3 sin  

sin 3  = 0,

we can conclude that the differential equation 

 = 3 sin( ) is not

separable.

3. 

 = 2

Solution:

Since



 = 2

 = 2 

1

2  =1



1

2  1

 = 0,

we can say that the differential equation 

 = 2 is not separable.

4. 

 = 1 + 2

Page 14 of 72

Solution:

Since



 = 1 + 2

 1 + 2  = 0,

we can conclude that the differential equation 

 = 1 + 2 is separable.

5. 

 + 4 = 8

Solution:

Since



 + 4 = 8



 = 8 4

 = (8 4 )

1

8 4 = 0

we can conclude that the differential equation 

 + 4 = 8 is separable.

6. 

 +  = 4

Solution:



 +  = 4



 = 4 



 = 4

1

4  = 

1

4  = 0,

we can conclude that the differential equation 

 +  = 4 is separable.

7. 

 + 4= 2

Solution:

Since



 + 4 = 2



 = 2 4

 = 2 4 

 2 4  = 0,

we can say that the differential equation 

 + 4= 2 is not separable.

8. 

 =  3 2 + 6

Page 15 of 72

Solution:

Since



 =  3 2 + 6



 = 2  3

1

3  2  = 0,

we can say that the differential equation 

 =  3 2 + 6 is separable.

9. 

 = sin +

Solution:

Since



 = sin +

 sin +  = 0,

we can say that the differential equation 

 = sin(+ ) is not separable.

10. 

 = 3 2

Solution:

Since



 = 3 2



 =

32

 32   = 0,

we can conclude that the differential equation 

 = 3 2 is separable.

***

Solution of Exercise 6 (General Solution of Separable D.E)

Find the general solution for each of the following. Where possible, write your answer

as an explicit solution.

1. 

 =  4

Solution:

Since 

 =  4 

 = 4 1

4   = 0, assume 4, we

obtain that

1

4   =1

ln 4 1

2 2 = 1

ln 4 =1

2 2 + 1

Page 16 of 72

4 = 1

2 2 + 1 , = 1

4 = 1

2 2

as the general solution for the differential equation 

 =  4 .

Note:

Since 4 =  1

2 2 ,

4 =  1

2 2  4

 1

2 2 = 1

2 2  < 4 =  1

2 2 + 4

 1

2 2 + 4

2. 

 = 3 2 2 sin

Solution:

Since 

 = 2 3sin  1

2  3 sin  = 0, we obtain that

1

2  3 sin   = 1

1

3 + cos  = 1

1

= 3 + cos + 1

= 1

3 + cos + 1

, =  1

=1

3 cos

as the general solution for the differential equation 

 = 3 2 2 sin .

3. 

 =  3 2 + 6

Solution:

Since 

 =  3 2 + 6 

 = 2 3 1

3  2  = 0,

we obtain

1

3  ( 2)  = 1

ln 3 1

2 2 + 2= 1

ln 3 =1

2 2 2+ 1

3 = 1

2 2 2+ 1 ,  = 1

3 = 1

2 2 2

as the general solution for differential equation 

 =  3 2 + 6.

Note:

Since 3 =  1

2 2 2 ,

Page 17 of 72

3 =  1

2 2 2  3

 1

2 2 2 = 1

2 2 2  < 3 =  1

2 2 2 + 3

 1

2 2 2 + 3

4. 

 = tan

Solution:

Since 

 = tan() 1

tan  = 0, assume tan() 0, we obtain

1

tan   = 1

cos

sin   = 1

1

sin( ) sin()  = 1

ln | sin | = 1

ln | sin | = + 1

sin  = + 1 ,  = 1

sin  =

as the general solution for 

 = tan( ).

Note:

Since sin  = ,

sin( ) =  0 sin() 1

 = 1 sin() < 0 = arcsin(  )

arcsin( ) 

5. 

 =

Solution:

Since 

 =

1

 1

 = 0, we obtain

1

 1

 = 1

ln ln = 1

ln =ln +ln 1

ln =ln( 1||) ,  = 1

= | |

as the general solution for the differential equation 

 =

.

6. 

 = 6 2 +4

3 2 4

Solution:

Since 

 = 6 2 +4

3 2 43 2 4  6 2 + 4 = 0, assume 3 2 4 0 we

obtain

3 2 4  6 2 + 4  =

3 2 2 2 3 4 =

Page 18 of 72

=2 3 + 4 +

2 2

as the general solution for the diffeential equation 

 = 6 2 +4

3 2 4 .

7. 2 + 1

 = 2 + 1

Solution:

Since 2 + 1

 = 2 + 1 1

1+ 2  1

1+ 2  = 0, we obtain

1

1 + 2  1

1 + 2  =

 tan arctan =

arctan = arctan +

= tan(arctan + )

as the general solution for differential equation 2 + 1

 = 2 + 1.

8. 2 1

 = 4 2

Solution:

Since 2 1

 = 4 2 2 1

2  4 = 0, assume 0, we obtain

2 1

2  4  =

1 1

2  4  =

+1

2 2 =

2 + 1

= 2 2 +

as the general solution for the differential equation 2 1

 = 4 2 .

9. 

 = 

Solution:

Since 

 =  1

  = 0  = 0, we obtain

  =

=

= +

=ln +

as the general solution for differential equation 

 =  .

10. 

 =  + 1

Solution:

Since 

 =  + 1 1

 +1  = 0, we obtain

1

 + 1   = 1

Page 19 of 72

+ 1   =1

1

+ 1  = 1

ln + 1 =1

ln + 1 = + 1

= + 1 1 ,  = 1

=ln | 1|

as the general solution for differential equation 

 =  + 1.

11. 

 = 3 3

Solution:

Since 

 = 3 3 1

3  3 = 0, assume 0, we obtain

1

3  3  =1

1

2 2 3

2 2 = 1

1

2 = 3 2 2 1, = 2 1

2 =1

3 2

As the general solution for differential equation 

 = 3 3 .

12. 

 = 2+

2+

Solution:

Since 

 = 2+

2+ 2 +  2 +  = 0, we obtain

2 +  2 +  =2

2 +2

3 2 2

3 = 2

2 1 + 1

3 = 2 1 + 1

3 + 2

3 +

3 = 3 +

3 + 2

2, 1= 2

2

= 3 +

3 + 1

3 +

3

,  = 3 1

= 3 + +

3 +

as the general solution for the differential equation 

 = 2+

2+ .

Page 20 of 72

13. 

 3 2 2 = 3 2

Solution:

Since 

 3 2 2 = 3 2 

 3 2 2 1 = 0 1

2 1  3 2  = 0, we

obtain

1

2 1  3 2  =

1

2

1 1

2

+ 1  3 2  =1

1

2ln 1 1

2ln + 1 3 = 1

ln 1

+ 1 = 2( 3 + 1 )

1

+ 1 = 2 3 + , = 2 1

1 = + 1 23 +

as the general solution from the differential equation 

 3 2 2 = 3 2 .

14. 

 3 2 2 = 3 2

Solution:

Since 

 3 2 2 = 3 2 

 = 3 2 1 + 2 1

1+ 2  3 2  = 0, we obtain

1

1 + 2  3 2  =

arctan 3 =

= tan( 3 + )

as the general solution of differential equation 

 3 2 2 = 3 2 .

15. 

 =200 2 2

Solution:

Since 

 =200 2 2 1

200 2 2  =  1

200 2 2  = 0, assume

100, we obtain

1

200 2 2   = 2

1

2 100   = 2

1

100

2 +

1

200

100   =2

1

200 ln 1

200 ln 100 = 2

1

200 ln

100 =+ 2

Page 21 of 72

ln

100 = 200 + 2 , 1=200 2

100 = 200 + 1 ,  = 1

= 100  200

as the general solution for the differential equation 

 =200 2 2 .

Solution of Quiz

1. Solve + 1 cos  + sin + 1  = 0.

Solution:

From the differential equation

+ 1 cos  + sin + 1  = 0,

dividing both side by + 1 (sin + 1), with the assumption that

sin( ) 1 and we obviously 1, we obtain

cos

sin + 1  +

+ 1  = 0.

cos

sin + 1  +

+ 1  =1

1

sin + 1 sin  + 1

+ 1 = 1

ln sin + 1 +ln + 1 = 1()

ln sin + 1 = ln + 1 + 1

sin = ln +1+ 1 1,  = 1

= arcsin  ln +1 1 1

= arcsin( + 1 1 1)

as the general solution of the separable differential equation + 1 cos  +

sin + 1  = 0. It is enough for us if we say that (*) is the general solution

of given separable differential equation.

2. Solve + 2  + + 4  = 0, 3 =1.

Solution:

From + 2  + + 4  = 0, dividing both side by + 2 ( + 4), with the

assumption that 2 and 4, we obtain

1

+ 4  +

+ 2  = 0

1

+ 4  +

+ 2  =

1

+ 4  + + 2 2

+ 2  =

1

+ 4  + 1 2

+ 2  =

Page 22 of 72

ln + 4 + 2ln + 2 =

2ln + 2 = ln + 4 +

ln( ) ln + 2 2 = ln + 4 +

ln

+ 2 2 = ln + 4 +ln

ln

+ 2 2 =ln

+ 4

+ 2 2 =

+ 4 ( )

Afterwards, we apply the initial condition to (*) and we obtain

1

1 + 2 2 =

3 + 4 =1.

Therefore, we can conclude that the solution of the initial- value problem under

consideration is

+ 2 2 = 1

+ 4

or it can be written as

+ 4

+ 2 2 = 1 .

Solution of Exercise 7 (Homogeneous D.E)

Definition

An equation in differential form ,  + ,  = 0 is said to be homogeneous,

if when written in derivative form 

 = , =

there exist a function such that

, =

.

Identify whether the following differential equations is homogeneous or not.

1.  + = 0

Solution:

Since  +  = 0 can be written as



 =

= 1 +

=

+

we obtain that there exist a function such that , =

. Therefore, we

can conclude that the differential equation  +  = 0 is

homogeneous.

2. 2 + 2  = 0

Page 23 of 72

Solution:

Since 2 +  2  = 0 can be written as



 = 2 + 

2 =

2

+

=1

2 +1

,

we obtain that there exist a function such that , =

. Thus, we can

conclude that the differential equation 2 +  2  = 0 is homogeneous.

3. 2 + + 2  2 +  = 0

Solution:

Since 2 +  + 2  2 +  = 0 can not be written as a function

,

that is



 = 2 +  + 2

2 + = 

+ 1 +



+1

=

+ 1 +

+1

can not be written as a function

, we say that the differential equation

2 +  + 2  2 +  = 0 is not homogeneous.

4. 2 + + 2  3 + 2  = 0

Solution:

Since 2 +  + 2  3 + 2  = 0 can not be written as a function

, that is,



 = 2 +  + 2

3 + 2 = (

+1+

)

2 2

2 + 1 =1

1

+ 1 +

2

+ 1

can not be written as a function

, we can conclude that the differential

equation 2 +  + 2  3 + 2  = 0 is not homogeneous.

***

Solution of Exercise 8 (Solution of Homogeneous D.E)

Theorem

If ,  + ,  = 0 is a homogeneous differential equation, then the change of

variables =  transform ,  + ,  = 0 into a separable equation in the

variables and .

Page 24 of 72

1. Find the general solution of 

 =  + 2

2.

Solution:

The given differential equation in derivative form is homogeneous since



 =  + 2

2 =

+

2

=

.

Now, let = . We obtain  =  +  

 =+ 

 and

+ 

 = + 2



 = 2

1

2  1

 = 0

integrating

1

2  1

 =

1

ln =

substituting =

, we obtain

ln = =

ln + ,

as the general solution of of 

 =  + 2

2.

2. Solve 2 

 = 2+ 2 given that = 0 at = 1.

Solution:

The given equation in derivative form is



 = 2 + 2

2 =

2 +

2 = 1

2

+1

2

and from this form we obtain that the given differential equation is

homogeneous. Now, let = . We obtain  =  +  

 =+ 

 and

+ 

 = 1

2 + 1

2



 = 1

2 1

2



 =1 2

2

2

1 2  1

 = 0

integrating

2

1 2  1

 = 1

1

1 2  2 1

 = 1

ln 12 ln = 1

Page 25 of 72

ln  12 = 1

ln  1 2 = (let = 1 )

 1 2 = .

Substituting =

we obtain

1 2 = 1 2

2 = 2

= .

Using the initial condition = 0 at = 1, we obtain

1 0 = 1 = . It means that

2

= 1. From this result, we obtain

2

= 1 2 = 2 + or 2

= 1 2 = 2 .

as the solution for the given initial value problem.

3. Solve 

 = +

and find the particular solution when 1 = 1.

Solution:

Since 

 = +

= 1 +

=

, +  = 0 is homogeneous.

Moreover, let =  and we will obtain that  =  +  

 =+ 



and

+ 

 = 1 + 

 = 1  1

 = 0 ( ).

Integrating (*), we obtain

 1

 = ln | | = 

Substituting =

to (**) we obtain = ln + .

Using initial condition, we obtain 1 = 1 ln 1 += 1.

Therefore, we can conclude that = ln | | + is the particular solution for the

given initial value problem.

4. Solve 

 = and find the particular solution when 2 = 1

2.

Since 

 = 

= 1

=

, = 0 is homogeneous.

Moreover, let =  and we will obtain that  =  +  

 =+ 



and

+ 

 = 1 

 = 1 2 1

1 2 1

 = 0 ( ).

Integrating (*), we obtain

1

1 2 1

 = 1

1

2ln 12 ln = 1

ln 1 2 +ln 2 = (let = 2 1)

ln  1 2 2 =

1 2 2 =

Page 26 of 72

2 2  2 = 

Substituting =

to (**) we obtain 2 2 = .

Using initial condition, we obtain 4 2 = 2 = .

Therefore, we can conclude that 2 2  = 2 is the particular solution for the

given initial value problem.

5. Solve 

 = 2

and find the particular solution when 1 = 1.

Solution:

Since 

 = 2

= 1 2

=

, 2  = 0 is homogeneous.

Moreover, let =  and we will obtain that  =  +  

 =+ 



and

+ 

 = 1 2 

 = 1 3 1

1 3 1

 = 0 ( ).

Integrating (*), we obtain

1

1 3 1

 = 1

1

3ln 13 ln = 1

ln 1 3 +ln( 2 ) = (let = 3 1)

ln  1 3  2 =

1 3 | | 2 =

3 3  3 = 

Substituting =

to (**) we obtain

3 3 2 = 2 3 = 3 =

2 .

Using initial condition, we obtain 1 + 3 =

1 = 4.

Therefore, we can conclude that 3 = 4

2= 2

2

is the particular solution

for the given initial value problem.

6. Given that 

 = +

 , prove that arctan

=1

2ln( 2 + 2 ) + where is an

arbitrary constant.

Proof:

Since



 = +

=

+ 1

1=

1

+ 1

1

1

=

, +  () = 0

is homogeneous. Moreover, let =  and we will obtain that

 = + 

 =+ 

 and

Page 27 of 72

+ 

 =

1

+ 1

1

1 

 = 1 + + 2

1 1

1 + 2  1

 = 0 ( ).

Integrating (*), we obtain

1

1 + 2  1

 =

1

1 + 2  1

2 2

1 + 2  1

 =

arctan 1

2ln 1 + 2 1

2ln( 2 ) =

arctan =1

2ln  1 + 2 2 + ()

Substituting =

to (*), we obtain

arctan( ) = 1

2ln 1 + 2

2 2 + =1

2ln( 2 + 2 ) + .

7. Find the general solution of 2 2 

 = 2+ 2.

Solution:

Since 

 = 2 + 2

2 2 = 1

2+ 1

2

2

=

, 2 + 2  2 2  = 0 is

homogeneous. Moreover, let =  and we will obtain that

 = + 

 =+ 

 and

+ 

 = 1 + 2

2 

 = 1 2

2 2

1 2  1

 = 0 ( ).

Integrating (*), we obtain

2

1 2  1

 = 2

1 ln | | = 2

1 =ln | | + 

Substituting =

to (**) we obtain

2

=ln + = 2

ln +

as the general solution of 2 2 

 = 2+ 2.

8. Find the general solution of 2 

 = 2 .

Solution:

Since



 =2

2 =

2

2

1

=

2 1

2

1

=

, (2 )  (2 )  = 0

is homogeneous. Moreover, let =  and we will obtain that

 = + 

 =+ 

 and

Page 28 of 72

+ 

 =

2 1

2

1 

 =2 1(2 )

2 2

2 1  1

 = 0 ( ).

Integrating (*), we obtain

2

2 1  1

 = 1

1

2 2

2 1  + 2

2 1  1

 = 1

1

2 1

2 1 2 + 1

1 1

+ 1  1

 = 1

1

2ln 2 1 +ln 1 ln + 1 ln = 1

ln 2 1 ln 1 2 +ln + 1 2 +ln( 2 ) = 2 (let 2=2 1)

ln 2 1  + 1 2 2

1 2 = 2

2 1  + 1 2 2

1 2 = 2 = (  ) let = 2 .

Substituting =

to (**), we obtain

2

2 1

+ 1 2 2

1 2 = 2 2

+ 1 2

2

2

= +

+ 2 =

as the general solution of 2 

 = 2 .

***

Solution of Exercise 9 (Non Homogeneous D.E)

Note: If

=

, suppose

=

=, then the differential equation

 + +  +  + +  = 0 can be written as

 + +  +  + +  = 0. To solve this, let =  + .

Find the solution of + + 1  + 2 + 2 + 1  = 0.

Solution:

Since + + 1  + 2 + 2 + 1  = 0 + + 1  + 2 + + 1  = 0 ,

let = + . We obtain  =  +  =  and

+ 1  + 2 + 1  = 0

+ 1   + 2 + 1  = 0

+ 1  + 1  + 2 + 1  = 0

+ 1  +  1 + 2 + 1  = 0

+ 1  + = 0.

Dividing both sides of + 1  +  = 0 by , we obtain

Page 29 of 72

+ 1

 + = 0.

Furthermore, we obtain

+ 1

 +  =

1 + 1

 +  =

+ln + = ()

Substituting = + to (*), we obtain

+ +ln + + =

+ 2 +ln | + | =

as the general solution of + + 1  + 2 + 2 + 1  = 0.

***

Solution of Exercise 10 (Solution of Non Homogeneous D.E)

1. Solve 3 + 1  6 2 3 = 0.

Solution:

Since

3 + 1  6 2 3  = 0 3 + 1  23 3  = 0,

let = 3 . We obtain  = 3 = 1

3 + 1

3 and

+ 1  2 3  = 0

+ 1 1

3 + 1

3 2 3 = 0

1

3 + 1  + 1

3 + 1  2 3 = 0

1

3 + 1

3 + 5

3 + 10

3 = 0.

Dividing both sides of 1

3+ 1

3 + 5

3+ 10

3 = 0 by 5

3+ 10

3, we obtain

+ 1

5 +10  + = 0.

Furthermore, we obtain

+ 1

5 +10  +  =2

5 + 1

5 5 +10  +  =2

1

5 5+ 10

5 +10 + 15

5 +10  +  =2

1

5 3

55

5 +10  +  =2

1

5 3

5 1

5 +10 5 +10 +  =2

1

5 3

5ln 5+ 10 + = 2()

Page 30 of 72

Substituting = 3 to (*), we obtain

1

5 3 3

5ln 5 3 + 10 + = 2

3 + 3 ln 15 + 5 +10 5 = 5 2

3 6 + 3 ln 5 3 + + 2  = 1

2 +ln 5 + ln 3 + + 2 = 1

3

2 +ln 3 + + 2 = 1

3 ln 5

2 +ln | 3 + + 2| =

as the general solution of 3 + 1  6 2 3  = 0.

2. Solve the initial value problem of 2 + 3 + 1  + 4 + 6 + 1  = 0;

2 = 2.

Solution:

Since

2 + 3+ 1  + 4 + 6+ 1  = 0

2 + 3 + 1  + 2 2 + 3 + 1  = 0,

let = 2 + 3 . We obtain  = 2 + 3 = 1

2 3

2 and

+ 1  + 2 + 1  = 0

+ 1 1

2 3

2 + 2+ 1  = 0

+ 1

2 3

2 3

2 + 2 + = 0

+ 1

2 + 1

2 = 0

Dividing both sides of +1

2 + 1

2 = 0 by 1

2 , we obtain

+ 1

1  + = 0

1 + 2

1  + = 0

1 + 2

1  + = 0.

Furthermore, we obtain

1 + 2

1  +  =1

+ 2 ln | 1| + = 1 ( )

Substituting = 2 + 3 to (*), we obtain

2 + 3 + 2 ln 2 + 3 1 + = 1

2 + 4 + 2 ln 2 + 3 1 =1

+ 2 +ln 2 + 3 1 = 1

2

+ 2 +ln |2 + 3 1| = ()

Afterwards, we substituting = 2 and = 2 (initial condition) to (**), we obtain

2 + 4 + ln | 4 + 6 1| = 2 + ln (1) = 2 = .

Page 31 of 72

Therefore, + 2 +ln |2 + 3 1| = 2 is the solution for the given initial value

problem.

***

Part 3

Solve the following differential equations.

1. 

 = 2 7

38

Solution:



 =2 7

3 8 = 27



3 8 1

 =1

2 7

3 8

Let =  =

. We obtain,  = + 

 =+ 

 and as a

consequence, we obtain

+ 

 = 1

2 7

3 8

+ 

 =2 7

3 8



 = 3 2 + + 2

3 8

3 8

3 2 2 1

 = 0

6

3 + 2 1

1  1

 = 0

2.3

3 + 2 1

1  +1

 = 0.

Furthermore, we obtain

2.3

3 + 2 1

1  + 1

 =

2ln 3 + 2 ln 1 +ln =

2ln 3

+ 2 ln

1 +ln | | =

as the general solusion of 

 = 2 7

38 .

2. 3 + 1

 + 6 2 = 6 2

Solution:

3 + 1 

 + 6 2 = 6 2

3 + 1 

 = 6 2 1

1

1  6 2

3 + 1  = 0.

Page 32 of 72

Afterwards, we obtain

1

1  2 3 2

3 + 1  = 1

1

1  2 1

3 + 1 3 = 1

ln 1 2ln 3 + 1 = 1

ln 1 +ln 3 + 1 2 = 1=

as the general solution of 3 + 1

 + 6 2 = 6 2 .

3. 

 = 2 2 + 2

2 2

Solution:



 =2 2 + 2

2 2 = 2 2 +

2

2 2

2 =1

2

2 +

2

2

1

2

.

Let =  =

. We obtain,  = + 

 =+ 

 and as a

consequence, we obtain

+ 

 = 1

2 2 + 2

2

1

2

+ 

 = 2 + 2

2 1



 = ( 2 2)

2 1

1 2

2 2  1

 = 0

1

2 1

+ 1  1

 = 0

Moreover,

1

2 1

+ 1  1

 = 1

ln | 2| ln | + 1| ln | | = 1

ln | 2| + ln | + 1| + ln | | = 1

ln  2  + 1 =

ln

2

+ 1 =

ln 2

2

2 =

ln 2

2 =

as the general solution of 

 = 2 2 + 2

2 2 .

4. 3 5  + +  = 0

Page 33 of 72

Solution:

3 5  + +  = 0



 = 3 5

+



 =

3 1

5

1 +

.

Let =  =

. We obtain,  = + 

 =+ 

 and as a

consequence, we obtain

+ 

 =  3

5

1 +

+ 

 = 3 + 5

1 +



 =  2 + 4 3

1 +

1 +

2 4 + 3  1

 = 0

2

3 1

1  1

 = 0

1

1 2

3  1

 = 0.

Furthermore, we obtain

1

1 2

3  1

 =

ln 1 2ln 3 ln =

ln

1 2ln

3 ln =

as the general solution of 3 5  + +  = 0.

5. 2 + 2  2 = 0; 1 = 0

Solution:

2 + 2 2 = 0 

 = 2 + 2

2 = 1

2. 1

+1

2.

.

Let =  =

. We obtain,  = + 

 =+ 

 and as a

consequence, we obtain

+ 

 = 1

2 +

2

+ 

 = 1 + 2

2



 =1 2

2

Page 34 of 72

2

1 2  1

 = 0.

Furthermore, we obtain

2

1 2  1

 = 1

1

1 2  2 1

 = 1

ln 12 ln = 1

ln 1 2 +ln =

ln 1 2

2 +ln | | = ( )

Substituting = 1 and = 0 (initial condition), we obtain

ln 1 0 + ln 1 = 0 = .

Therefore, ln 1 2

2 +ln | | = 0 is the solution of the given inital value problem.

6. 5 + 2+ 1  + 2 + + 1  = 0

Solution:

Let = 5 + 2 + 1 and = 2 + + 1. We obtain  = 5 + 2 and

 = 2 +. Using elimination, we obtain  = 2 and  = 5 2.

Afterwards, we obtain

 + = 0

 2 + 5  2 = 0

 2 + 5 2 = 0

2  + 2 + 5  = 0

1 2

 +2 + 5

 = 0.

Let =

=. As a consequence, we obtain  = + and

1 2  + 2 + 5  + = 0

 2 2 2 + 5 2  + 5 = 0

1 4 + 5 2  + 2 + 5  = 0

1

 + 2 + 5

1 4 + 5 2  = 0.

Moreover, we obtain

1

 + 2 + 5

1 4 + 5 2  =

1

 +1

2 4 + 10

1 4 + 5 2  =

1

 +1

2 1

1 4 + 5 2 14 + 5 2 =

ln +1

2ln 14 + 5 2 =

ln |5 + 2 + 1| + 1

2ln 1 4 2+ + 1

5 + 2+ 1 + 5 2+ + 1

5 + 2 + 1 2 =

as the general solution of 5 + 2+ 1  + 2 + + 1  = 0.

Page 35 of 72

Solution of Exercise 11 (Exact and Non Exact D.E)

( In exercises 1-10 determine whether or not each of the given equation is exact; solve

those that are exact).

1. 3 + 2  + 2 +  = 0.

Solution:

Our first duty is to determine whether or not the equation is exact or not. Here

, = 3 + 2 , , = 2 + ,

 ,

 = 2,  ,

 = 2.

Since  ,

 =  ,

 = 2

we can conclude that the differential equation 3 + 2  + 2 +  = 0 is

exact differential equation. Furthermore, we must find such that

 ,

 = , = 3 + 2 and  ,

 = , = 2 + .

From the first of these,

, = ,  + = 3 + 2  + =3

2 2 + 2 + .

Then  ,

 = 2 + 



But we must have  ,

 = , = 2 + .

Thus

2 + = 2 + 

 = 

 .

Thus = 1

2 2 + 0 , where 0 is an arbitrary constant, and so

, =3

2 2 + 2 +1

2 2 + 0 .

Hence a one- parameter family of solution is , = 1, or

3

2 2 + 2 +1

2 2 + 0 = 1

Combining the constsnts 0 and 1 we may write this solution as

3

2 2 + 2 +1

2 2 =

where = 1 0 is an arbitrary constant.

So, we conclude that the general solution of the exact differential equation

3 + 2  + 2 +  = 0 is 3

2 2 + 2 + 1

2 2 =.

2. ( 2 + 3) + 2 4 = 0.

Solution:

Our first duty is to determine whether or not the equation is exact or not. Here

, = 2 + 3, , = 2  4,

Page 36 of 72

 ,

 = 2 ,  ,

 = 2 .

Since  ,

 =  ,

 = 2

we can conclude that the differential equation ( 2 + 3) + 2  4  = 0 is

exact differential equation. Afterwards, we must find such that

 ,

 = , = 2 + 3 and  ,

 = , = 2 4.

From the first of these,

, = ,  + = 2 + 3  + = 2 + 3 + .

Then  ,

 = 2  + 



But we must have  ,

 = , = 2 4.

Thus

2 4 = 2 + 

 4 = 

 .

Thus = 4 + 0, where 0 is an arbitrary constant, and so

, =  2 + 3 + 4 + 0 .

Hence a one- parameter family of solution is , = 1, or

 2 + 3 + 4 + 0 = 1

Combining the constsnts 0 and 1 we may write this solution as

 2 + 3 + 4 =

where = 1 0 is an arbitrary constant.

So, we conclude that the general solution of the exact differential equation

( 2 + 3) + 2 4  = 0 is 2 + 3 + 4 = .

3. 2 + 1  + 2 + 4  = 0.

Solution:

Our first duty is to determine whether or not the equation is exact or not. Here

, = 2  + 1, , = 2 + 4 ,

 ,

 = 2 ,  ,

 = 2 .

Since  ,

 =  ,

 = 2

we can conclude that the differential equation 2  + 1  + 2 + 4  = 0 is

exact differential equation. Afterwards, we must find such that

 ,

 = , = 2 + 1 and  ,

 = , = 2 + 4.

From the first of these,

Page 37 of 72

, = ,  + = 2  + 1  + = 2 + + .

Then  ,

 = 2 + 



But we must have  ,

 = , = 2 + 4 .

Thus 2 + 4= 2 + 

 4 = 

 .

Thus = 2 2 + 0 , where 0 is an arbitrary constant, and so

, = 2 + + 2 2 + 0 .

Hence a one- parameter family of solution is , = 1, or

2 + + 2 2 + 0 = 1

Combining the constsnts 0 and 1 we may write this solution as

2 + + 2 2 =

where = 1 0 is an arbitrary constant.

So, we conclude that the general solution of the exact differential equation

2  + 1  + 2 + 4  = 0 is 2 + + 2 2 = .

4. 3 2 + 2  3 +  = 0.

Solution:

Our first duty is to determine whether or not the equation is exact or not. Here

, = 3 2 , , =  3,

 ,

 = 3 2 ,  ,

 = 3 2 .

Since  ,

 = 3 2  ,

 = 3 2

we can conclude that the differential equation 3 2 + 2  3 +  = 0 is

not exact (non- exact) differential equation.

5. 6 + 2 2 5 + 3 2 + 4 6 = 0.

Solution:

Our first duty is to determine whether or not the equation is exact or not. Here

, = 6  + 2 2 5, , = 3 2 + 4  6,

 ,

 = 6 + 4 ,  ,

 = 6 + 4 .

Since  ,

 =  ,

 = 6 + 4

we can conclude that the differential equation

6  + 2 2 5  + 3 2 + 4  6  = 0 is exact differential equation.

Moreover, we must find such that

 ,

 = , = 6 + 2 2 5 and  ,

 = , = 3 2 + 4 6.

Page 38 of 72

From the first of these,

, = ,  +

= 6 + 2 2 5 +

= 3 2 + 2 2 5 +

Then  ,

 = 3 2 + 4  + 



But we must have  ,

 = , = 3 2 + 4 6.

Thus

3 2 + 4 6 = 3 2 + 4 + 

 6 = 

 .

Thus = 6 + 0, where 0 is an arbitrary constant, and so

, = 3 2 + 2  2 5 6 + 0 .

Hence a one- parameter family of solution is , = 1, or

3 2 + 2 2 5 6 + 0 = 1

Combining the constsnts 0 and 1 we may write this solution as

3 2 + 2 2 5 6 =

where = 1 0 is an arbitrary constant.

So, we conclude that the general solution of the exact differential equation

6  + 2 2 5  + 3 2 + 4  6  = 0 is 3 2 + 2  2 5 6 = .

6. 2 + 1 cos  + 2 sin  = 0.

Solution:

Our first duty is to determine whether or not the equation is exact or not. Here

, = 2 + 1 cos , , = 2 sin ,

 ,

 = 2 cos ,  ,

 = 2 cos .

Since  ,

 =  ,

 = 2 cos

we can conclude that the differential equation 2 + 1 cos  + 2 sin  = 0

is exact differential equation. Moreover, we must find such that

 ,

 = , = 2 + 1 cos and  ,

 = , = 2 sin

From the first of these,

, = ,  +

= 2 + 1 cos  +

= 2 + 1 sin +

Then

Page 39 of 72

 ,

 = 2 sin + 



But we must have  ,

 = , = 2 sin .

Thus

2 sin = 2 sin + 

 0 = 

 .

Thus = 0, where 0 is an arbitrary constant, and so

, = 2 + 1 sin + 0 .

Hence a one- parameter family of solution is , = 1, or

2 + 1 sin + 0 = 1

Combining the constsnts 0 and 1 we may write this solution as

2 + 1 sin =

where = 1 0 is an arbitrary constant.

So, we conclude that the general solution of the exact differential equation

2 + 1 cos  + 2 sin  = 0 is 2 + 1 sin = .

7. sec2 + sec tan  + tan + 2  = 0.

Solution:

Our first duty is to determine whether or not the equation is exact or not. Here

, = sec2 + sec tan , , = tan + 2 ,

 ,

 = sec2 ,  ,

 = sec2 .

Since  ,

 =  ,

 = sec2

we can conclude that the equation sec2 + sec tan  + tan + 2  =

0 is exact differential equation. Furthermore, we must find such that

 ,

 = , = sec2 + sec tan and  ,

 = , = tan + 2.

From the first of these,

, = ,  +

= sec2 + sec tan  +

= tan + sec + .

Then  ,

 = tan + 



But we must have  ,

 = , = tan + 2 .

Thus

Page 40 of 72

tan + 2 = tan + 

 2 = 

 .

Thus = 2 + 0, where 0 is an arbitrary constant, and so

, = tan + sec + 2 + 0.

Hence a one- parameter family of solution is , = 1, or

tan + sec + 2 + 0 = 1

Combining the constsnts 0 and 1 we may write this solution as

tan + sec + 2 =

where = 1 0 is an arbitrary constant.

So, we conclude that the general solution of the exact differential equation

sec2 + sec tan  + tan + 2  = 0 is tan + sec + 2 = .

8.

2+ + 2

3+ = 0.

Solution:

Our first duty is to determine whether or not the equation is exact or not. Here

, =

2 + ,, = 2

3 + ,

 ,

 = 2

3 ,  ,

 =2

3 .

Since  ,

 = 2

3  ,

 =2

3

we can conclude that the differential equation

2+ + 2

3+ = 0 is

not exact (non- exact) differential equation.

9. 21

 +  2

2  = 0.

Solution:

Our first duty is to determine whether or not the equation is exact or not. Here

, =2 1

, , = 2

2 ,

 ,

 =1 2

2 ,  ,

 =1 2

2 .

Since  ,

 =  ,

 =1 2

2

we can conclude that the differential equation 21

 +  2

2  = 0 is exact

differential equation. Furthermore, we must find such that

 ,

 = , =2 1

and  ,

 = , = 2

2 .

From the first of these,

, = ,  +

=2 1

 +

Page 41 of 72

= 2

+ .

Then  ,

 = 2

2 +

2 + 

 = 2

2 + 



But we must have  ,

 = , = 2

2 .

Thus 2

2 = 2

2 + 

 0 = 

 .

Thus = 0, where 0 is an arbitrary constant, and so

, = 2

+ 0 .

Hence a one- parameter family of solution is , = 1, or

2

+ 0 = 1

Combining the constsnts 0 and 1 we may write this solution as

2

=

or we can write this as 2 = , where = 1 0 is an arbitrary constant.

So, we conclude that the general solution of the exact differential equation

21

 + 2

2  = 0 is 2 =.

10. 23

2+1

1

2 + 3 1

2 1

21 = 0.

Solution:

Our first duty is to determine whether or not the equation is exact or not. Here

, =2 3

2+ 1

1

2

, , = 3 1

2 1

21,

 ,

 = 3 1

2

1

2

, ,

 = 3

21

2 1

2=3

2 1

2

1

2

.

Since  ,

 = 3 1

2

1

2,

 = 3

2 1

2

1

2

we can conclude that the differential equation 2 3

2+1

1

2 + 3 1

2 1

21 = 0

is not exact (non- exact) differential equation.

Page 42 of 72

Solution of Exercise 12 (Integrating Factor)

( In each of the following equations determine the constant such that the equation is

exact, and solve the resulting exact equation).

a. 2 + 3 + 2+ 4  = 0.

Solution:

Suppose , = 2 + 3 and , =  2 + 4 .

Then, we obtain  ,

 = 3 and  ,

 = 2 .

In order to make the differential equation become exact differential equation, it

must be  ,

 =  ,

 = 3 .

Thus

2 = 3 = 3

2.

Therefore, we obtain that the differential equation

2 + 3  + 3

2 2 + 4 = 0 is exact differential equation. Furthermore,

we must find such that

 ,

 = , = 2 + 3 and  ,

 = , =3

2 2 + 4.

From the first of these,

, = ,  + = 2 + 3  + =1

3 3 + 3

2 2 + .

Then  ,

 = 3

2 2 + 

 .

But we must have  ,

 = , =3

2 2 + 4.

Thus

3

2 2 + 4=3

2 2 + 

 4 = 

 .

Thus = 2 2 + 0 , where 0 is an arbitrary constant, and so

, =1

3 3 + 3

2 2 + 0 .

Hence a one- parameter family of solution is , = 1, or

1

3 3 + 3

2 2 + 2 2 + 0 = 1

Combining the constsnts 0 and 1 we may write this solution as

1

3 3 + 3

2 2 + 2 2 =

where = 1 0 is an arbitrary constant.

Page 43 of 72

So, we conclude that the general solution of the exact differential equation

2 + 3  + 3

2 2 + 4 = 0 is 1

3 3 + 3

2 2 + 2 2 =.

b. 1

2+ 1

2  +  +1

3  = 0.

Solution:

Suppose , = 1

2 +1

2 and , =  + 1

3 .

Then, we obtain  ,

 = 2

3 and  ,

 =

3 .

In order to make the differential equation become exact differential equation, it

must be  ,

 =  ,

 = 2

3 .

Thus

3 = 2

3 = 2.

Therefore, we obtain that the differential equation

1

2+ 1

2  + 2 +1

3  = 0 is exact differential equation. Furthermore, we

must find such that

 ,

 = , =1

2 +1

2 and  ,

 = , = 2 + 1

3 .

From the first of these,

, = ,  + = 1

2 +1

2  + = 1

+

2 + .

Then  ,

 = 2

3 + 

 .

But we must have  ,

 = , = 2 + 1

3 .

Thus 2 + 1

3 = 2

3 + 

 1

3 = 

 .

Thus = 1

2 2 + 0 , where 0 is an arbitrary constant, and so

, = 1

+

2 1

2 2 + 0 .

Hence a one- parameter family of solution is , = 1, or

1

+

2 1

2 2 + 0 = 1

Combining the constsnts 0 and 1 we may write this solution as

Page 44 of 72

1

+

2 1

2 2 =

where = 1 0 is an arbitrary constant.

So, we conclude that the general solution of the exact differential equation

1

2+ 1

2  + 2 +1

3  = 0 is 1

+

2 1

2 2 =.

Solution of Exercise 13 (Grouping Method)

( Solve using grouping method)

1. 1



2  = 0.

Solution:

From 1



2  = 0 we obtain

= .

So

=

is the general solution of the differential equation 1



2  = 0.

2. 2 

 + 2 2 = 0.

Solution:

2 

 + 2 2 = 0

2  

 = 2 2

2 2  2 = 0

From 2 2  2  = 0, we group the term as follows

2 ( 2  + 2 ) = 0.

Thus 2  2 = .

So, 2  =

is the general solution of the differential equation 2 

 + 2 2 = 0.

3. 2 + 1  + 2 2  = 0.

Solution:

From 2 + 1  + 2 2  = 0, we group the term as follows

2   + 2  + 2  4 = 0.

Thus 2  + 2 2 2 = .

Page 45 of 72

Therefore,

2 + 2 2 2 =

is the general solution of the differential equation

2 + 1  + 2 2  = 0.

4. 2 + 6  + 2 + 4 3  = 0.

Solution:

From 2  + 6  + 2 + 4 3  = 0, we group the term as follows

2  + 2  + 6  + 4 3  = 0.

Thus 2 + 3 2 + 4 = .

So, 2 + 3 2 + 4 =

is the general solution of the differential equation

2  + 6  + 2 + 4 3 = 0.

Solution of Quiz

1. Which of the following differential equations can be made exact by multiplying

by 2?

(a) 

 + 2

= 4.

Solution: 

 + 2

= 4 

 = 4 2

4 2

 = 0

By multiplying both sides by 2, we obtain

4 2 2  +  2  = 0.

Here , = 4 2 2  , , =  2 ,

 ,

 = 2 ,  ,

 = 2

Since

 ,

 =  ,

 = 2

we can conclude that the differential equation 3 + 2  + 2 +  = 0

is exact differential equation. In the other word, the differential equation



 + 2

= 4 can be made exact by multiplying by 2.

(b) 

 + 3= 2 .

Solution:



 + 3 = 2 

 = 2 3 2 3  +   = 0.

By multiplying both sides by 2, we obtain

4 3 2  +  3 = 0.

Page 46 of 72

Here , = 4 3 2 , , = 3 ,

 ,

 = 3 2 ,  ,

 = 3 2 .

Since  ,

 =  ,

 = 3 2

we can conclude that the differential equation

4 3 2  +  3  = 0 is exact differential equation. In the other

word, the differential equation 

 + 3= 2 can be made exact by

multiplying by 2.

(c) 1



 1

2 = .

Solution:

1



 1

2 = 1



 = +1

2  +1

2  + 1

 = 0.

By multiplying both sides by 2, we obtain

3 +  +   = 0.

Here , = 3 +, , =  ,

 ,

 = 1,  ,

 = 1.

Since  ,

 = 1  ,

 = 1

we can conclude that the differential equation

3 +  +  = 0 is not exact differential equation. In the other

word, the differential equation 1



 1

2 = can not be made exact by

multiplying by 2.

(d) 1



 + 1

2 = 3.

Solution:

1



 + 1

2 = 3 1



 = 3 1

2 3 1

2  + 1

 = 0.

By multiplying both sides by 2, we obtain

3 2  +   = 0.

Here , = 3 2 , , = ,

 ,

 = 1,  ,

 = 1.

Since  ,

 =  ,

 = 1

we can conclude that the differential equation

Page 47 of 72

3 2  +   = 0 is exact differential equation. In the other word,

the differential equation 1



 + 1

2 = 3 can be made exact by multiplying by

2 .

2. Consider the differential equation

4 + 3 2  + 2 = 0.

(a) Show that this equation is not exact.

Proof. Here , = 4 + 3 2 , , = 2 ,

 ,

 = 6 ,  ,

 = 2 .

Since  ,

 = 6  ,

 = 2

we can conclude that the differential equation 4 + 3 2  + 2 = 0 is



(b) Find an integrating factor of the form , where n is a positive integer.

Solution:

From (a) we know that the differential equation 4 + 3 2  + 2 = 0

is not exact. But then, we can find an integrating factor , = , where

is a positive integer such that the differential equation 4 + 3 2  +

2  = 0 is exact.

Assume that 4 + 3 2  + 2 = 0 is exact differential equation.

Here

4 + 3 2  + 2  = 0

4 +1 + 3 2  + 2 +1  = 0.

Let , = 4 +1 + 3 2 and , = 2 +1 .

Then, we obtain

( , )

 = 6 and  ,

 = 2 + 2

Since 4 + 3 2  + 2  = 0 is exact differential equation, we must

obtain

6 = (, )

 =  ,

 = 2 + 2 ,

that is

6 = 2 + 2 .

Hence = 2.

Thus, the integrating factor of the form , where is a positive integer such

that the differential equation 4 + 3 2  + 2  = 0 is exact is 2.

(c) Multiplying the given equation through by the integrating factor found in (b)

and solve the resulting exact equation.

Solution:

Page 48 of 72

From (b) we know that the differential equation 4 3 + 3 2 2  +

2 3  = 0 is exact. Now, we will find the solution of this exact differential

equation or in the other word we must find such that

 ,

 = , = 4 3 + 3 2 2 and  ,

 = , = 2 3 .

From the first of these,

, = , +

= 4 3 + 3 2 2  +

= 4 + 3 2 + .

Then  ,

 = 2 3 + 



But we must have  ,

 = , = 2 3 .

Thus

2 3 = 2 3 + 

 0 = 

 .

Thus = 0, where 0 is an arbitrary constant, and so

, = 4 +3 2 + 0 .

Hence a one- parameter family of solution is , = 1, or

4 +3 2 + 0 = 1

Combining the constsnts 0 and 1 we may write this solution as

4 +3 2 =

where = 1 0 is an arbitrary constant.

So, we conclude that the general solution of the exact differential equation

4 3 + 3 2 2 + 2 3  = 0 is 4 + 3 2 = .

Solution of Exercise 14 (Linear D.E)

Solve the given differential equations.

1. 

 + 3

= 6 2.

Solution:

Here = 3

, = 6 2

and hence  = 3

 = 3ln = ln 3= 3.

Therefore, we obtain

 =  

Page 49 of 72

3 = 3  6 2 

3 = 6 +

= 3 +1

3

as the genaral solution of the given differential equation, where is an arbitrary

constant.

2. 4

 + 2 3 = 1.

Solution: 4 

 + 2 3 = 1 

 +2 3

4 =1

4 

 + 2

=1

4 .

Here = 2

, =1

4

and hence  = 2

 = 2ln = ln 2= 2.

Therefore, we obtain

 =  

2 = 2 1

4 

2 = 1

+

= 1

3 +1

2

as the genaral solution of the given differential equation, where is an arbitrary

constant.

3. 

 + 3 = 3 2 3 .

Solution:

Here = 3, = 3 2 3

and hence  = 3 = 3 .

Therefore, we obtain

 =  

3 = 3 (3 2 3 )

3 = 3 +

= 3

3 +1

3

as the genaral solution of the given differential equation, where is an arbitrary

constant.

4. 

 + 4 = 8 .

Solution:

Page 50 of 72

Here = 4, = 8

and hence  = 4 = 2 2 .

Therefore, we obtain

 =  

22 =  22 (8 ) 

22 = 2  22 (4 ) 

22 = 2 22 2 2

22 = 2 22 +

= 2 + 1

22

as the genaral solution of the given differential equation, where is an arbitrary

constant.

Solution of Exercise 15 (Solution of Linear D.E)

Consider the differential equation 

 + = 0.

(a) Show that if and are two solutions of this equation and 1 and 2 are arbitrary

constants, then 1 + 2 is also a solution of this equation.

Proof:

Here 

 + = 0 

 =  

+  = 0 .

By integrating both sides of (*), we obtain



+  =0

ln =  + 0

| | =  0

=   ,

where 0 and are constants, as the general solution of the differential equati on



 + = 0. Now, let and are two solutions of that equation, and are

given as follow =  and =  ,

where and are constants.

For 1 and 2 are arbitrary constants, we have

1 + 2 = 1  + 2 

= 1 + 2 

Page 51 of 72

=  (),

where is constant. From (*) we can conclude that 1 + 2 also a solution of the

differential equation 

 + = 0 .

(b) Extending the result of (a), show that if 1, 2,, are solutions of this equation

and 1 , 2 , , are arbitrary constants, then

=1

is also a solution of this equation.

Proof:

By extending the result of (a), let 1, 2,, are solutions of the differential

equation 

 + = 0. Then

1= 1  , 2= 2  ,, =  ,

where 1, 2,, and are constants.

For 1 , 2 , , are arbitrary constants, we obtain

=1

= 1 1  + 2 2  + + 

= 1 1+ 2 2++ 

=  ( ),

where is constant. From (*) we can conclude that

=1 is also a solution of

the differential equation



 + = 0 

Solution of Exercise 16 (Properties of Linear D.E)

(a) Let 1 be a solution of 

 + = 1

and 2 be a solution of 

 + = 2 ,

Where , 1, and 2 are all defined on the same real interval . Prove that 1+ 2 is

a solution of 

 + = 1 + 2

On .

Proof:

Since 1 is a solution of 

 + = 1

and 2 is a solution of 

 + = 2 ,

we have

Page 52 of 72

1= 1=  1 

 and 2= 2=  2 

 .

Since , 1, and 2 are all defined on the same real interval , we know that the

solution of



 + = 1 + 2 ,

is

=  1 + 2 



=  1() +  2() 



=  1 

 +  1 



= 1+ 2.

(b) Use the result of (a) to solve the equation



 + = 2 sin + 5 sin 2 .

Solution:

Let = 1 , 1 = 2 sin , and 2 = 5 sin 2 .

Since

1 =  1 



= 2 sin 

= sin cos + 1

= sin cos + 1

is the solution of 

 += 2 sin ,

and

2 =  2 



= 5 sin 2

= sin 2 2 cos 2 + 2

= sin 2 2 cos 2 + 2

is the solution of 

 += 2 sin ,

using the result (a), we obtain

= 1 + 2

= sin cos + 1

+ sin 2 2 cos 2 + 2

= sin cos + sin 2 2 cos 2 + 1 + 2

= sin cos + sin 2 2 cos 2 +  ; = 1+ 2

as the solution of

Page 53 of 72



 + = 2 sin + 5 sin 2 .

Notes:

2 sin  = 2 sin 2 cos 

= 2 sin 2 cos + 2 sin 

=2 sin 2 cos + 0

2

= sin cos + 1 .

5 sin 2 = 5 sin 2 2 5 cos 2

= 5 sin 2 10 cos 2 + 4 5 sin 2

=5 sin 2 10 cos 2 + 0

5

= sin 2 2 cos 2 + 2 .

Solution of Exercise 17 (Integrating Factor of Linear D.E)

Solve each differential equation by first finding an integrating factor.

1. 5 + 4 2 + 1  + 2 + 2 = 0.

Solution:

Here , = 5 + 4 2 + 1 and , = 2 + 2

Since

 ,

 = 5  + 4 2 + 1

 = 5 + 8  ,

 = 2 + 2 

 = 2 + 2 ,

we can say that the differential equation 5  + 4 2 + 1  + 2 + 2  = 0

is not exact (non- exact) differential equation. Furthermore, we will find an

integrating factor ( , ) such that the differential equation

,  5  + 4 2 + 1  + ( , ) 2 + 2  = 0 is exact differential

equation. Let ( ) is the integrating factor depends only upon . Then we have

=  ,

  ,



,  .

Since  ,

  ,



, =5 + 8 2 2

2 + 2  = 3( + 2 )

( + 2 )= 3

,

we obtain = 3

 = ln 3= 3.

Multiplying the differential equation by this integrating factor, we obtain exact

differential equation

5 4 + 4 3 2 + 3  + 5 + 2 4  = 0.

Page 54 of 72

Moreover, we will find a solution of this exact differential equation by using

grouping method. From 5 4 + 4 3 2 + 3  + 5 + 2 4  = 0, we group

the terms as follows

5 4  + 5  + 4 3 2  + 2 4  + 3  = 0.

Thus 5 + 4 2 + 1

4 4 = .

So, 5 + 4 2 + 1

4 4 =

is the general solution of exact differential equation 5 4 + 4 3 2 + 3  +

5 + 2 4  = 0.

2. 2 + tan  + 2 tan  = 0.

Solution:

Here , = 2 + tan and , = 2 tan .

Since

 ,

 = 2 + tan

 = sec2 ,

 = 2 tan

 = 1 2 tan ,

we can say that the differential equation 2 + tan  + 2 tan  = 0

is not exact (non- exact) differential equation. Furthermore, we will find an

integrating factor ( , ) such that the differential equation

,  2 + tan  + ( , ) 2 tan  = 0 is exact differential

equation. Let ( ) is the integrating factor depends only upon . Then we have

=  ,

  ,



,  .

Since

 ,

  ,



, =sec2 1 + 2 tan

2 + tan = tan tan + 2

2 + tan = tan ,

we obtain

= tan  =sin

cos  = 1

cos cos = ln |cos | = cos .

Multiplying the differential equation by this integrating factor, we obtain exact

differential equation

2 cos + sin  + cos 2 sin  = 0.

Moreover, we will find a solution of this exact differential equation by using

grouping method. From 2 cos + sin  + cos 2 sin  = 0, we

group the terms as follows

2 cos  2 sin  + sin  + cos  = 0.

Thus 2 cos + sin = .

So, 2 cos + sin =

is the general solution of the exact differential equation 2 cos + sin  +

cos 2 sin  = 0.

3. 2 + 1 +  + 2 + 1  = 0.

Solution:

Here

Page 55 of 72

, = 2 + 1 + =  2 + 2 + and , = 2  + 1.

Since

 ,

 = (  2 + 2 + )

 = 2  + 2+ 1  ,

 = (2  + 1)

 = 2 ,

we can say that the differential equation 2 + 1 +  + 2  + 1  = 0 is

not exact (non- exact) differential equation. Furthermore, we will find an

integrating factor ( , ) such that the differential equation

,  2 + 1 +  + ( , ) 2  + 1  = 0 is exact differential

equation. Let ( ) is the integrating factor depends only upon . Then we have

=  ,

  ,



,  .

Since  ,

  ,



, =2  + 2 + 1 2

2 + 1 = 1,

we obtain = = .

Multiplying the differential equation by this integrating factor, we obtain exact

differential equation

[ 2 ( + 1) +  ] + 2 +  = 0.

Moreover, we will find a solution of this exact differential equation by using

grouping method. From [ 2 ( + 1) +  ]  + 2  +  = 0, we

group the terms as follows

[ 2 + 1  + 2  ] + (  +  ) = 0.

Thus  2 +  = .

So,  2 +  =

is the general solution of exact differential equation [ 2 ( + 1) +  ]  +

2  +  = 0.

4. 2 2 +  + 2 3  = 0.

Solution:

Here , = 2 2 + dan , = 2 3 .

Since  ,

 = (2  2 + )

 = 4  + 1  ,

 = (2 3 )

 = 1,

we can say that the differential equation 2  2 +  + 2 3  = 0 is not

exact (non- exact) differential equation. Furthermore, we will find an integrating

factor ( , ) such that the differential equation

,  2  2 +  + ( , ) 2 3  = 0 is exact differential equation. Let

( ) is the integrating factor depends only upon . Then we have

=  ,

  ,



,  .

Since  ,

  ,



, =4  + 2

2 2 + = 2 2 + 1

2  + 1 = 2

,

Page 56 of 72

we obtain = 2

 = 2ln | |= ln 1

2 =1

2 .

Multiplying the differential equation by this integrating factor, we obtain exact

differential equation 2 + 1

 +2

2  = 0.

Moreover, we will find a solution of this exact differential equation by using

grouping method. From 2 + 1

 + 2

2  = 0, we group the terms as

follows

2 + 1



2  + 2 = 0.

Thus 2 +

+ 2 = .

So, 2 +

+ 2 =

is the general solution of the exact differential equation

2 +1

 +2

2  = 0.

Find an integrating factor of the form and solve.

4  2 + 6  + 5 2 + 8  = 0.

Solution:

Here , = 4 2 + 6 and , = 5 2 + 8.

Since

 ,

 = (4  2 + 6 )

 = 8  + 6  ,

 = 5 2 + 8

 =10  + 8,

we can say that the differential equation 4  2 + 6  + 5 2 + 8  = 0 is not

exact (non- exact) differential equation. Furthermore, we will find an integrating factor

of the form such that the differential equation

4 +1 +2 + 6 +1  + 5 +2 +1 + 8 +1  = 0

is exact differential equation.

Since 4 +1 +2 + 6 +1  + 5 +2 +1 + 8 +1  = 0

is exact differential equation,

4 +1 +2 + 6 +1

 = 5 +2 +1 + 8 +1



4 + 2 +1 +1 + 6 + 1 = 5 + 2 +1 +1 + 8 + 1 .

From (*) we have

4 + 2 = 5 + 2 5 4 = 2 

and

6 + 1 = 8 + 1 8 6 = 2 4 3 = 1  .

Page 57 of 72

From (**) and (***) we obtain = 2 and = 3. Hence, 2 3 is the integrating factor as

desired. Moreover, we will find the solution of the exact differential equation

4 3 5 + 6 2 4 + 5 4 4 + 8 3 3 = 0

by using grouping method. From 4 3 5 + 6 2 4  + 5 4 4 + 8 3 3  = 0, we

group the terms as follows

4 3 5  + 5 4 4  + 6 2 4  + 8 3 3  = 0.

Thus 4 5 + 2 3 4 = .

So, 4 5 + 2 3 4 =

is the general solution of the exact differential equation

4 3 5 + 6 2 4  + 5 4 4 + 8 3 3 = 0.

Solution of Exercise 18 (Orthogonal and Oblique Trajectories)

In exercises 1 4 find the orthogonal trajectories of each give family of curves. In each

case sketch several numbers of the family and several of the orthogonal trajectories on

the same set of axes.

1. = 3 .

Solution:

Step 1.

We first find the differential equation of the given family

=  3 i .

Differentiating, we obtain 

 = 3  2 ii .

Eliminating the parameter between Equations (i) and (ii), we obtain the

differential equation of the family (i) in the form



 =3

iii .

Step 2

. We now find the differential equation of the orthogonal trajectories by

replacing 3 / in iii by its negative reciprocal, obtaining



 =

3 iv .

Step 3

. We now solve the differential equation iv . Separating variables, we have

3 = .

Integrating, we obtain the one- parameter family of solutions of i in the form

3

2 2 + 1

2 2 = 1

2 or 2+ 3 2= 2,

where is an arbitrary constant.

Page 58 of 72

2. 2 =.

Solution:

Step 1.

We first find the differential equation of the given family

2 =  i .

Differentiating, we obtain

2

 = ii .

Eliminating the parameter between Equations (i) and (ii), we obtain the

differential equation of the family (i) in the form



 =

2 iii .

Step 2

. We now find the differential equation of the orthogonal trajectories by

replacing /2 in iii by its negative reciprocal, obtaining



 = 2

iv .

Step 3

. We now solve the differential equation iv . Separating variables, we have

 = 2 .

Integrating, we obtain the one- parameter family of solutions of i in the form

1

2 2 + 2 = 1

2 or 2 2+ 2= 2,

where is an arbitrary constant.

Page 59 of 72

3.  2 + 2 = 1.

Step 1.

We first find the differential equation of the given family

 2 + 2 = 1 i .

Differentiating, we obtain

2 + 2

 = 0 or 

 = 

ii .

Eliminating the parameter between Equations (i) and (ii), we obtain the

differential equation of the family (i) in the form



 = 1 2

 iii .

Step 2

. We now find the differential equation of the orthogonal trajectories by

replacing (1 2 )/ in iii by its negative reciprocal, obtaining



 = 

1 2 iv .

Step 3

. We now solve the differential equation iv . Separating variables, we have

1

 = .

Integrating, we obtain the one- parameter family of solutions of i in the form

ln | | 1

2 2 1

2 2 = 1 or 2 + 2 ln( 2 ) =

where is an arbitrary constant.

Page 60 of 72

4. = .

Step 1.

We first find the differential equation of the given family

=  i .

Differentiating, we obtain 

 =  ii .

Eliminating the parameter between Equations (i) and (ii), we obtain the

differential equation of the family (i) in the form



 =ln

ln or 

 = ln

iii .

Step 2

. We now find the differential equation of the orthogonal trajectories by

replacing ln | | / in iii by its negative reciprocal, obtaining



 =

ln iv .

Step 3

. We now solve the differential equation iv . Separating variables, we have

ln | |  = .

Integrating, we obtain the one- parameter family of solutions of i in the form

1

2 2 ln | | 1

4 2 + 1

2 2 = 1 or 2 2 ln | | 2 + 2 2 = ,

where is an arbitrary constant.

Page 61 of 72

***

12.

Find the value of such that the parabolas = 1 2 + are the orthogonal trajectories

of the family of ellipses 2 + 2 2 = 2 .

Solution:

Step 1.

We first find the differential equation of the given family

2 + 2 2 =2 i .

Differentiating, we obtain

2 + 4

 

 = 0 or 

 = 2

4 1 ii .

Step 2

. We now find the differential equation of the orthogonal trajectories by replacing

2 /(4 1) in ii by its negative reciprocal, obtaining



 =4 1

2 iv .

Step 3

. We now solve the differential equation iv . Separating variables, we have

1

4 1 = 1

2 .

Integrating, we obtain the one- parameter family of solutions of i in the form

1

4ln |4 1| 1

2ln | | = 2 or = 1 2 + 1

4,

where 2 and 1 are arbitrary constants.

Hence, the value of that we want is 1/4, = 1/4.

***

13.

Find the value of such that the curves + = 1 are orthogonal trajectories of the

family =

1 2 .

Solution:

Page 62 of 72

Step 1.

We first find the differential equation of the given family

=

1 2 i .

Differentiating, we obtain 

 = 1

1 2 2 ii .

Eliminating the parameter 2 between Equations (i) and (ii), we obtain the differential

equation of the family (i) in the form



 = 1

1 1

 2 or 

 = 2

2 iii .

Step 2

. We now find the differential equation of the orthogonal trajectories by replacing

2 / 2 in iii by its negative reciprocal, obtaining



 = 2

2 iv .

Step 3

. We now solve the differential equation iv . Separating variables, we have

2  =2 .

Integrating, we obtain the one- parameter family of solutions of i in the form

1

3 3 + 1

3 3 = 2 or 3 + 3 = 1,

where 2 and 1 are arbitrary constant.

Therefore, the value of that we want is 3, = 3.

***

15.

Find a family of oblique trajectories that intersect the family of circles 2 + 2 = 2 at

angle 45o .

Solution:

Step 1

. We first find the differential equation of the given family

2 + 2 =2 .

Differentiating, we obtain

2 + 2

 = 0 or 

 =

i .

Step 2.

We replace , =  / in Equation (i) by

, + tan

1 , tan =

+ 1

1 +

=

+ .

Thus the differential equation of the desired oblique trajectories is



 =

+ ii .

Step 3

. We now solve the differential equation (ii). Observing that it is a homogeneous

differential equation, we let =  to obtain

+ 

 = 1

+ 1.

Page 63 of 72

After simplifications this becomes + 1

2 + 1  = 

.

Integrating we obtain

1

2ln 2 + 1 + arctan( ) = ln + 1

or

ln 2 2 + 1 + 2 arctan = .

Replacing by / , we obtain the family of oblique trajectories in the form

ln 2 + 2 + 2 arctan

= ,

where is an arbitrary constant.

***

16.

Find a family of oblique trajectories that intersect the family of parabolas 2 =  at

angle 60o .

Solution:

Step 1

. We first find the differential equation of the given family

2 =  i .

Differentiating, we obtain

2

 = or 

 = 1

2 ii .

Eliminating the parameter between Equation (i) and (ii), we obtain the differential

equation 

 =

2 iii .

of the given family of parabolas.

Step 2.

We replace , = /2 in Equation (iii) by

, + tan

1 , tan =

2 + 3

1 3

2 .

Thus the differential equation of the desired oblique trajectories is



 =

2 + 3

1 3

2 iv .

Step 3

. We now solve the differential equation (iv). Observing that it is a homogeneous

differential equation, we let =  to obtain

+ 

 =

1

2 + 3

1 1

2 3

= + 2 3

2 3.

After simplifications this becomes

Page 64 of 72

1

2 2 3 1 3

2

2 3 + 2 3  = 

.

Integrating

1

2 2 3 1

2 3 + 2 3  3

2 1

3

2

3 3 + 2  = 1



1

2 2 3 1

2 3 + 2 3  3

2 1

2 2 3

6 + 3

36 + 69

36  = 1



1

2 2 3 1

2 3 + 2 3  3

2 1

 3

6 2

+ 69

6 2  = 1



1

2 2 3 1

2 3 + 2 3  3

2

36

69 69

6 6

69

6 3

69 2

+ 1  = 1



1

2ln 2 + 2 3 3

23 3 arctan 6 3

69 +ln = 1

ln | 2 + 2 3| 6

23 3 arctan 6 3

69 +ln 2 = .

Substituting = / , we obtain

ln 2  2 2 3 6

23 3 arctan 6

3

69 = .

Therefore, the family of oblique trajectories in the form

ln 2  2 2 3 6

23 3 arctan 6

3

69 = .

***

Solution of Exercise 19 (Problem in Mechanics (Frictional Forces))

A man is pushing a loaded sled across a level field of ice at the constant speed of 10

ft/sec. When the man in halfway across the ice field, he stops pushing and lets the

loaded sled continue on. The combined weight of the sheld and its load is 80 lb

( =80 lb ); the air resistence (in pounds) is numerically equal to 3

4, where is the

velocity of the sheld (in feet per second); and the coefficient of fraction of the runners

on the ice is 0.04 ( = 0.04 ). How far will the sheld continue to move after the man

stops pushing?

Solution:

Page 65 of 72

0.04  80 3

4 = 5

2



16

5 3

4 = 5

2



1

5 = 1

32

5 3

2

1

5 = 1

3

2 32

5

Integrating

1

5 = 1

3

2 32

5

1

5 =2

3 1

3

2 32

5 3

2

1

5 = 2

3ln 3

2 32

5 + 1

3

10 + 2 =ln 3

2 32

5

3

10 +64

15 =( )

We know that 0 =10 feet/sec. Thus

+64

15 = 150

15 = 86

15.

Therefore

86

15 3

10 +64

15 = .

The shled will stop when = 0.

3

10 =32

43 10

3ln  32

43 =

Integrating this we obtain = 172

9 3

10 +64

15 + 4

We know that 0 = 0. Thus

172

9= 4 .

Therefore = 172

9 3

10 +64

15 + 172

9

and  = 10

3ln  32

43 9. 09 feet.

Therefore, in 9.09 feet the sheld will continue to move after the man stops pushing.

***

Page 66 of 72

Solution of Exercise 20 (Rate Problems (Rate of Growth and Decay and

Population) Growth)

5. Assume that the population of a certain city increases at a rate proportional to

the number of its inhabitants at any time. If the population doubles in 40 years,

in how many years will it triple?

Solution:

Let be the number of individuals in at time . We know that the population of a

certain city increases at a rate proportional to the number of its inhabitants at

any time. Hence, we are led to the differential equation



 =  i ,

where is a constant of proportionality. The population is positive and is

increasing and hence  /  > 0. Therefore, from (i), we must have > 0. Now,

suppose that at time 0= 0 the population is 0. Then, in addition to the

differential equation (i), we have the initial condition

0 = (0) = 0 ii .

The differential equation (i) is separable. Separating variables, integrating, and

simplifying, we obtain =   .

Applying the initial condition, = 0 at = 0= 0, to this, we have

0 =  0 =

From this we at once find = 0  0 and hence we obtain the unique solution

= 0  0

of the differential equation (i), which satisfies the initial condition (ii). Now,

when =40 , we have = 2 0 . Hence, we obtain

2 0 = 0 40 2 = 40 ln 2

40 =.

If we let = 3 0 , then we obtain

3 0 = 0 ln 2

40 3 = ln 2

40 ln 3 = ln 2

40  = 40 ln 3

ln 2 63.40

Therefore, the population will triple in about 63 .40 years.

6. The population of the city of Bingville increases at a rate proportional to the

numbers of its inhabitants present ant any time . If the population of Bingville

was 30 ,000 in 1970 and 35 ,000 in 1980 , what will be the population of Bingville

in 1990 ?

Solution:

According to the formula in the exercise 5, we have

= 0  0 .

Hence we obtain

1980 =30, 000 1980 1970 35,000

30, 000 = 10 1

10 ln 7

6 =.

Therefore, the population of Bingville in 1990 is

Page 67 of 72

1990 =30 , 000 2ln7

640,833.

***

9.

The human population of a certain island satisfies the logistic law



 =  2 (i)

with = 0.03 = 3 10 2 , = 3 108 , and time measured in years.

(a) If the population in 1980 is 200 ,000 , find a formula for the population in future

years.

Solution:

We must solve the separable differential equation (i) subject to the initial

solution 1980 =200, 000 ii .

Separating variable in (ii), we obtain



 = 0.03 3 10 8 2 

3102 3108 2 =

and hence 

3102 1 ( 106 ]=.

Using partial fractions, this becomes

100

3 1

+ 10 6

1 106 =.

Integrating, assuming 0 < <106 , we obtain

100

3 ln ln 1 10 6  = + 1

and hence

ln

1 106 = 3

100 + 2 .

Thus we find

1 106 = 3

100 .

Solving this for , we finally obtain

=  3

100

1 + 106  3

100 iii.

Now, applying the initial conditions (ii) to this, we have

210 5 = 59.4

1 + 10 6  59.4 ,

from which we obtain

=2 10 5

59.4 1 2 10 5 10 6 = 10 6

4 59.4 .

Substituting this value for back into (iii) and simplifying, we obtain the solution

in the form

Page 68 of 72

= 10 6

1 + 4 59.4 3

100 iv.

This gives the population as a function of time for >1980 .

Therefore, we can conclude that the formula for the population in future years is

given by = 10 6

1 + 4 59.4 3

100

.

(b) According to the formula of part (a), what will be the population in the year

2000?

Solution:

Let =2000 in (iv) and we obtain

= 10 6

1 + 40.6 312 ,965.

Therefore, the population in the year 2000 is 312 ,965 people.

(c) What is the limiting value of the population as ?

Solution:

lim

 = lim

 ? 10 6

1 + 4 59.4 3

100 = 10 6 = 1,000,000.

***

Solution of Exercise 21 (Mixture Problem)

18.

A large tank initially contains 200 gal of brine in which 15lb of salt is dissolved. Starting

at = 0, brine containing 4lb of salt per gallon flows into the tank at the rate of 3.5

gal/min. The mixture is kept uniform by stirring and the well- stirred mixture leaves the

tank at the rate of 4 gal/min.

(a) How much salt is in the tank at the end of one hour?

Solution:

Let denotes the amount of salt in the tank at time . We apply the basic

equation 

 = .

The brine flows in at the rate of 3.5 gal/min, and each gallon contains 4 lb of salt.

Thus  =(4 lb/gal)(3.5 gal/min)=14 lb/min,

and  = ( lb/gal)(4 gal/min)=4 lb/min,

where lb/gal denotes the concentration. But here, since the rate of outflow is

different from that of inflow, the concentration is not quite so simple. At time

= 0, the tank contains 200 gal of brine. Since brine flows in at the rate of 3.5

gal/min but flows out at the faster rate of 4 gal/min, there is a gross gain of

Page 69 of 72

3.5 4 = 0.5 gal/min of brine in the tank. Thus at the end of minutes the

amount of brine in the tank is 200 0.5 gal.

Hence the concentration at time minutes is

200 0.5 lb/gal,

and so  = 4

200 0.5 lb/min.

Thus the differential equation becomes



 =14 4

200 0.5 = 14 8

400 i .

Since there was initially 15 lb of salt in the tank, we have the initial condition

0 =15 ii .

The differential equation (i) is not separable but it is linear. Putting it in standard

form, we obtain 

 + 8

400 = 14,

we find the integrating factor

exp 8

400  = 400 8 =1

400 8 .

Multiplying through by this we have

1

400 8 

 + 8

400 9 =14

400 8

or

 1

400 8  =14

400 8 .

Thus

1

400 8 =2

400 7 +

or = 400 + 400 8 .

Applying condition (ii), =15 at = 0, we obtain

15 = 400 + 4008

or = 375

4008 .

Thus the amount of salt at any time > 0 is given by

= 400 375 400

400 8

Therefore, the amount of salt in the tank at the end of one hour ( =60 ) is

60 =340 375 340

400 8

=340 375 238 lb .

(b) How much salt is in the tank when the tank contains only 50 gal of brine?

Page 70 of 72

Solution:

According to part (a), we know that at the end of minutes the amount of brine

in the tank is 200 0.5 gal.

Thus we obtain

200 0.5 =50

and hence =300.

Therefore, the amount of salt in the tank when the tank contains only 50 gal of

brine is 300 =100 375 100

400 8 100 lb.

Solution of Exercise 22 (Reduction of Order)

1. Prove that the differential equation

2

 2 

 2 = 0

has two solutions of the form = , with is a constant.

Proof:

Let = be the solution of the given differential equation. Then we obtain

2 

 2 

 2  = 0

 

   2  = 0

2   2  = 0

2 2 = 0

+ 1  2 = 0

= 0  + 1 = 0 2 = 0.

Hence there are two values for , i.e., = 1 or = 2. Therefore, we can conclude

that the given differential equation has two solutions of the form = , i.e.,

=  or = 2 .

2. If a differential equation has general solution of the form

= 1 2 + 2 3 ,

then determine that differential equation.

Solution:

Consider the second order homogeneous differential equation with constant

coefficient 2

 2 + 

 +  = 0.

Suppose that = , is the solution of that differential equation (where will

be determined). Then we will have the characteristic equation

2 +  + = 0.

Page 71 of 72

Thus, in order to = becomes the solution for that differential equation, has to

satisfies that characteristic equation. One of the possibilities for the roots of the

characteristic equation above is if the characteristic equation has two different roots,

say 1, 2, then the general solution for the differential equation above is

= 1 1 +2 2 .

Since we have that the general solution of a differential equation is

= 1 2 + 2 3

we obtai that the roots of a characteristic equation are 1= 2 or 2=3. Moreover,

the characteristic equation that we want is

2 + 3 = 0 2 + 6 = 0.

Hence we obtain = 1 and = 6.

Therefore, the diferential equation that has the general solution

= 1 2 + 2 3

is 2

 2 + 

 6 = 0.

3. Given that 1 = 2 is one of the solutions of the differential equation

2

 2 + 1



 4

2 = 0 (i)

Determine a general solution of that differential equation.

Solution:

Observe that 1 = 2 does satisfy the given differential equation, i.e.,

2 2

 2 + 1

2

 4

2 2 = 2 + 1

2 4 = 0.

Let 2 ( ) = 2 ( ). We have

2

 = 2

 + 2 

and

2 2

 2 = 2 2

 2 + 4

 + 2 .

Substituting the expression for ,  / , and 2 /  2 into Equation (i), we obtain

2 2

 2 + 4

 + 2 + 1

 2

 + 2  4

2 2 = 0

or 2 2

 2 + 5

 = 0.

Letting =  /  we obtain the first- order homogeneous linear equation

2 

 + 5  = 0.

Treating this as a separable equation, we obtain

1

 = 5

1

5 + 1

 = 0.

Integrating this, we obtain the general solution

1

5ln | | + ln | | = 6 ln  1

5 =ln 5 = 5

5

5 = 4

5

where 6 , 5, and 4 are constants.

Page 72 of 72

Afterwards, we choose 4= 1, we recall that  /  = and integrate to obtain the

function given by = 1

4 4 .

Now forming 2 = 1 ( ), where 1() denotes the known solution, we obtain

the function 2 defined by

2 = 2  1

4 4 = 1

4 2 = 1

42 .

Therefore, the general solution of Equation (i) is

( ) = 1 2 + 3 1

42 = 1 2+ 2 2

where 1 , 2, and 3 are constants.

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