(module 2) MAT235 Chapter 4 page 231-245 (answer for separable equation)

MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

CHAPTER 4
ORDINARY DIFFERENTIAL EQUATION (ODE)

List of Topics

 First Order ODE
 3 methods to solve the first order ODE
i) Separable equation
ii) Homogeneous equation
iii) Linear equation

 Applications of First Order ODE
i) Population Growth and Bacterial Colony
ii) Radioactive Decay
iii) Newton’s Law of Cooling
iv) Mixing Problem
v) Electrical Circuits (Kirchhoff’s Law)

 Second Order ODE
i) Homogeneous equation
ii) Non-homogeneous equation

4.0 Introduction

Definition of Degree

i) dy = x + y ………………. degree ____
dx y degree ____
degree ____
ii) dy = x 2 + y 2 …………… degree ____
dx xy degree ____

iii) dy = xy + y 2 ……………
dx xy

v) dy = x 2y + xy 2 ……………
dx y 3

v) dy = x 2y + y 3 ……………
dx xy 2

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

Definition of Order (highest derivative)

i) dy = x 2 + 1 ………………….. order ____
dx

ii) d 2y + dy + y = sin x ………. order ____
dx 2 dx

iii) d 4 y + dy + y = e x ………. order ____
dx 4 dx

iv) d2y +  dy 3 + y = ln x …… order ____
dx 2  dx 

4.1 First Order ODE
4.1.1 3 methods to solve the first order ODE

4.1.1.1 Separable equation

Procedure to solve separable equation
Step I : separating variables x and y

LHS : f(y) dy
RHS : f(x) dx

Step II : Integrating both sides (using an appropriate technique)

LHS : f (y ) dy
RHS : f (x) dx

(if the initial condition is given)
Step III : Applying the given initial condition to find the value of C.

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

Example 1 Ans : ey = − e−x + C
Solve ex +y dx − e2x +2y dy = 0
Solution :

ex +y dx − e2x +2y dy = 0

I) Separating variables x and y, LHS : f (y )dy and LHS : f (x )dx

e x+y dx = e 2x+2y dy

Or
e2x +2y dy = ex +y dx

e2x • e2y dy = ex • ey dx

e2y dy = ex dx
ey e2x

e2y −y dy = ex −2x dx

ey dy = e−x dx

 II) Integrating both sides, LHS : f (y )dy and LHS : f (x)dx

 ey dy = e−x dx

ey = e−x + C
−1

ey = − e−x + C #

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

Example 2

( ) ( )Solve y 2 1+ x 3 y' + x 2 1+ y 2 = 0 ans : y − tan−1 y = − 1 ln 1+ x3 + C
3

Solution :

( ) ( )y 2 1+ x 3 y' + x 2 1+ y 2 = 0

( ) ( )y 2 1+ x3 dy + x2 1+ y 2 = 0
dx

I) Separating variables x and y, LHS : f (y )dy and LHS : f (x )dx

( ) ( )y 2 1+ x3 dy = − x2 1+ y 2
dx

( ) ( )y 2 1+ x3 dy = − x2 1+ y 2 dx

y 2 2 dy = − x2 dx
1+ 1+ x3
y

 II) Integrating both sides, LHS : f (y )dy and LHS : f (x)dx

 y 2 dy = − x2 dx --------------- 
1+ 1+ x3
y2

LHS : RHS :

By doing long division Using u-substitution

==

= = (into )
=
= =
Hence, =

= 1 dx = tan−1 x + C
= 1+ x2

( )because d 1
(into ) dx tan−1 x = 1+ x2

Thus, the solution is, 234
y − tan−1 y = − 1 ln1+ x3 + C #
3

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

Example 3

Solve xy + y' e x2 ln y = 0 ans : 1 (ln y )2 = 1 e−x2 + C

22

Solution :

xy + y' e x2 ln y = 0

xy + dy ex 2 ln y = 0
dx

I) Separating variables x and y, LHS : f (y )dy and LHS : f (x )dx

dy ex 2 ln y = − xy
dx

ln y dy = −x dx
y ex2

ln y dy = − xe−x 2 dx
y

 II) Integrating both sides, LHS : f (y )dy and LHS : f (x)dx

 ln y dy = − xe−x2 dx ------------- 
y

LHS : RHS :

Using u-substitution (into ) Using u-substitution (into )
= =
= =
= =
= =
= =

Thus, the solution is,
1 ln2 y = 1 e− x 2 + C #
22

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

Example 4 – with initial condition

Solve 2tan x dy + 1 = 0 , when x=, y =2 ans : ln y + 1 = − ln sin x + ln 3
y 2 − 1 dx 2 y −1

Solution :

2tan x dy + 1 = 0
y 2 − 1 dx

I) Separating variables x and y, LHS : f (y )dy and LHS : f (x )dx

2 tan x dy = −1
y 2 − 1 dx

2 dy = − 1 dx
y2 −1 tan x

2 dy = − cot x dx
y2 −1

 II) Integrating both sides, LHS : f (y )dy and LHS : f (x)dx

 y2 1 dy = − cot x dx ------------- 
2−

LHS : ------- (i)
Using partial fraction

=

 ------- (ii)
When
→

When →

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

LHS :

Substitute and into (i)

=
= (into )

RHS :

Change to basic trigonometry and use u-substitution,
=
=
=
=
=
= (into )

Therefore, the general solution is,
ln y − 1 − ln y + 1 = − ln sin x + C ------------------ 

III) Applying the given initial condition into  (to find C)

By substituting x =  , y = 2 into ,
2

ln 2 − 1 − ln 2 + 1 = − ln sin   + C Thus, the particular solution is,
2

ln1 − ln 3 = − ln(1 )+ C

0 − ln 3 = 0 + C ln y − 1 − ln y + 1 = −ln sin x − ln 3 #

C = − ln 3 (into )

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

Example 5 (exercise 1/page 55/i) – with initial condition

( )Solve ln y cos 2 x − 1 dy − cos x dx = 0 , when x =  , y = 1
2
ans : y ln y − y = cosec x − 2
Solution :

( )ln y cos 2 x − 1 dy − cos x dx = 0

I) Separating variables x and y, LHS : f (y )dy and LHS : f (x )dx

( )ln y cos2 x − 1 dy = cos x dx

cos x
cos2 x −
( )ln y dy = 1 dx

cos x
1− cos2
( )ln y dy = − x dx

ln y dy = cos x dx
− sin2 x

ln y dy = − cos x dx

(sin x)2

 II) Integrating both sides, LHS : f (y )dy and LHS : f (x)dx

 ln y dy = − cos x dx ------------- 

(sin x)2

LHS : (integration by parts) RHS : (integration using u- substitution)

= =
= =
= =
= =
= (into ) =
= (into )

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

Therefore, the general solution is,
y ln y − y = 1 + C ------------------ 
sin x

III) Applying the given initial condition into  (to find C)

By substituting x =  , y = 1 into ,
2

(1)ln(1) − 1= 1 +C
sin 
2

−1= 1 +C
1

−1=1 +C

C = − 2 (into )

Thus, the particular solution is,
y ln y − y = 1 − 2 #

sin x
or y ln y − y = cos ecx − 2

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

More Examples from Previous Semester Papers

 June 2019/ MAT235/ Q4b/ 7 marks

Solve the separable differential equation dy = 2ex where y (0) = 1 .
dx y 2
2

Solution : ∶ 1 3 = 2 − 47
3 24

Step I : Separating variables x and y
2
= 2
2 = 2

Step II : Integrating both sides

∫ 2 = ∫ 2

3 = 2 +
3

Step III : initial condition = , =



3 = 2 +
3

1 3 = 2 +
3

1 13 = 2 0 +
3 (2)

11
3 (8) = 2 +

1
= 24 − 2

47
= − 24

 1 3 = 2 − 47
3 24

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

 June 2018/ MAT235/ Q4b/ 7 marks

Solve the separable differential equation dy − 2xey = ey where y (0) = 1.

dx

Solution : ∶ − − = + 2 − 1


Step I : Separating variables x and y

− 2 =


= + 2


= (1 + 2 )


1
= (1 + 2 )

− = (1 + 2 )

Step II : Integrating both sides

∫ − = ∫(1 + 2 )

− − = + 2 2 +
2

− − = + 2 +

Step III : initial condition = , =

− −1 = 0 + 02 +

1
= −

 − − = + 2 − 1 #


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MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

 Jan 2018/ MAT235/ Q4a/ 14 marks ∶ ) 3 − 7 = 1 −1 5 +
Solve the following first order differential equations : 3 4 (4)

( )i) 25x2 + 16 dy = 5 dx . (solution in page 242)
y2 −7

(Hint : dx = 1 tan−1 x  + C ) (7 marks)
x2 + a2 a a
(7 marks) (solution in page 266)
ii) dy = 2y 2 − 2x2 ∶ ) − |( )2 + 1| = +
dx 4xy

Solution :

Step I : Separating variables x and y

(25 2 + 16) = 2 5 7
−

( 2 − 7) = 5 16
25 2 +

Step II : Integrating both sides

∫( 2 − 7) = ∫ 5 1265)
( 2 +
25

∫( 2 − 7) = 1 ∫ 1 16
5 + 25
2

∫( 2 − 7) = 1 ∫ 1
5 + (54)2
2

3 − 7 = 1 ∙ 1 −1 +
3 5 ( )

3 − 7 = 1 ∙ 1 −1 +
3 5 (54) ((54))

3 − 7 = 1 ∙ 5 −1 5 +
3 5 4 (4)

3 − 7 = 1 −1 5 +
3 4 (4)

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

 Mar 2016/ MAT235/ Q4b/ 8 marks

Solve the separable differential equation dy = x cos x where y (0) = − 1.
Solution : dx y
2 1
∶ 2 = + − 2

Step I : Separating variables x and y ∶ ∫

= Using Tabular Integration dv (integrate)
Sign u (differentiate)
= +
Step II : Integrating both sides −
−1
∫ = ∫ +0
2
2 = + + ∫ = + − ∫ 0

Step III : initial condition = , = − ∫ = + +
(−1)2
2 = 0 0 + 0 +
1
2 = 0 + 1 +
1
= − 2
2 1
 2 = + − 2 #

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

 Sep 2015/ MAT235/ Q4a(i)/ 4 marks

Solve the first order differential equation dy = 3x2 given that y (0) = 1.
Solution : dx y

∶ 2 = 3 + 1
2 2

Step I : Separating variables x and y
3 2
=

= 3 2
Step II : Integrating both sides

∫ = ∫ 3 2

2 3 3
2 = 3 +

2 = 3 +
2

Step III : initial condition = , =

(1)2 = (0)3 +
2

1
= 2

 2 = 3 + 1 #
2 2

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 4 : Ordinary Differential Equations (ODE)

 Mar 2015/ MAT235/ Q4a/ 12 marks ∶ ) = 1 2 + 1
Solve the following first order differential equations : 2 2

i) dy = xex2 −y , given that y (1) = 1. (5 marks) (Solution in page 245)

dx

ii) dy = cos x − y tan x , given that y (0) = 1. (7 marks) (Solution in page 281)

dx
∶ ) = +

Solution :

Step I : Separating variables x and y

= 2−


2
=

= 2 ∶ ∫ 2

Step II : Integrating both sides

∫ = ∫ 2 ∫ 2 = ∫ ∙ = 2
2
= 2
1 2 = 1 ∫
2 2
= + = 2
1
= 2 +

= 1 2 +
2
Step III : initial condition = , =

(1) = 1 (1)2 +
2

1
= 2 +

1
= 2

 = 1 2 + 1 #
2 2

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