(module 2) MAT235 Chapter 4 page 246-269 (answer for homogeneous equation)

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

4.1.1.2 Homogeneous equation

Procedure to solve Homogeneous equation

Method I :

Step I : write dy as a subject
dx

( )Step II : write dy into the form ofy and name as equation 
dx x

(Note : if the expression has the same degree, dividing each expression by x deg ree )

Step III : let u = y
x

Therefore, y = ux

using product rule,

dy = x du + u ---------------- 
dx dx

Step IV : substitute  into  and separating variables x and u
LHS : f(u) du
RHS : f(x) dx

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

LHS : f (u) du
RHS : f (x) dx

Step VI : Replace u by y
x

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

© Amirah 2022 246

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

Procedure to solve Homogeneous equation

Method II :

Step I : write as a subject


Step II : Substitute = + and =



Step III : factorize and simplify x

Step IV : separating variables x and v

LHS : f(v) dv

RHS : f(x) dx

Step V : Integrating both sides (using an appropriate technique)
LHS : ∫ ( )
RHS : ∫ ( )

Step VI : Replace v by y
x

(if the initial condition is given)

Step VII : Applying the given initial condition to find the value of C.

© Amirah 2022 247

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

Example 1 ( ) ( )ans : ln2
y +1 + tan−1 y = − 2ln x + C
Solve (x + 2y )dy = (y − 2x)dx x x

Solution : Method I

Step I : write dy as a subject Homogeneous equation : write as a subject.
dx

(x + 2y )dy = (y − 2x)dx

dy = y − 2x (homogeneous equation with degree 1)
dx x + 2y

( ) ( )Step II : write dy into the form ofy u= y and name as equation 
dx x x
, replace

(Note : if the expression has the same degree, dividing each expression by x deg ree )

y − 2x
x
( )dy y
= x (dividing each term by x1) → to write into the form of x

dx x x + 2y x

( )dy =y −2
x
( )dx 1 + 2 y x

dy = u − 2 Step III :
dx 1+ 2u
dy = u − 2 ----------------  Where
dx 1+ 2u and

Step IV : substitute  into  and separating variables x and u Differentiate (using product rule)
----------------- 
 =  : x du + u = u − 2
dx 1+ 2u

x du = u − 2 − u
dx 1+ 2u

x du = u − 2 − u(1+ 2u)

dx 1+ 2u

x du = u − 2 − u − 2u2
dx 1+ 2u

x du = − 2 − 2u2
dx 1+ 2u

( )x du = − 2 u2 + 1
dx 2u + 1

2u + 1du = − 2 dx
u2 +1 x

© Amirah 2022 248

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

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

+ 1du
+1
 2u = − 2 dx
u2 x

  u 2u 1 du + u 1 1 du = −2 1 dx
2+ 2+ x

ln u2 + 1 + tan−1(u) = − 2ln x + C

Step VI : Replace u by y
x

( ) ( )ln
y 2 +1 + tan−1 y x = − 2ln x + C #
x

Solution : Method II

Step I : write as a subject


( + 2 ) = ( − 2 )
− 2
= + 2

Step II : Substitute = + and = ( )



( ) − 2
+ = + 2( )

− 2
+ = + 2

Step III : factorize and simplify
( − 2)

+ = (1 + 2 )
− 2

+ = 1 + 2

© Amirah 2022 249

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

Step IV : separating variables x and v Step V : Integrating both sides (using an
appropriate technique)

− 2 1 + 2 −2
= 1 + 2 −

− 2 − (1 + 2 ) ∫ 1 + 2 = ∫
=
1 + 2 1 2 1

− 2 − − 2 2 ∫ (1 + 2 + 1 + 2) = −2 ∫

= 1 + 2 1 2 1
∫ 1 + 2 + ∫ 1 + 2 = −2 ∫
−2 − 2 2
= 1 + 2 −1( ) + |1 + 2| = −2 +

−2(1 + 2) Step VI : Replace v by ( )
= 1 + 2 −1 ( ) + |1 + ( )2| = −2 + #

1 + 2 −2
1 + 2 =

Example 2

dy = y 2 − x 2 ( ) ( )ans : 1 ln2 +1
dx x 2 + 2xy + y 2 2
Solve y + tan−1 y x = −ln x + C
x

Solution : Method I

Step I : write dy as a subject
dx

dy = y 2 − x 2 (homogeneous equation with degree 2)
dx x 2 + 2xy + y 2

( ) ( )Step II : write dy into the form ofy u= y and name as equation 
dx x x
, replace

(Note : if the expression has the same degree, dividing each expression by x deg ree )

dy y2 − x2 ( )(dividing each term by x2) → to write into the form ofy
dx x2 x2 x
=
x 2 + 2xy + y 2
x 2 xx x 2

( )dy= y2 −1
( ) ( )dx x
1+ + y 2
2 yx x

© Amirah 2022 250

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

dy = u2 − 1
dx 1 + 2u + u2

dy = u2 − 1
dx u2 + 2u + 1

dy = (u + 1)(u −1) Step III :
dx (u + 1)(u + 1)
Where
dy = u − 1 ----------------  and
dx u + 1
Differentiate (using product rule)
Step IV : substitute  into  and separating variables x and u ----------------- 

 =  : x du + u = u − 1
dx u + 1

x du = u − 1 − u
dx u + 1

x du = u −1− u(u + 1)

dx u + 1
x du = u −1− u2 − u

dx u + 1

x du = − u2 − 1
dx u + 1

( )x du = − u2 + 1
dx u + 1

uu2++11du =− 1 dx
x

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

 uu2++11du = − 1 dx
x

u  u1 du + u 1 du = − 1 dx
2+ 2+ 1 x

u  12u du + u 1 1 du = − 1 dx
2+ 1 2+ x
2

1 ln u2 + 1 + tan−1(u) = − ln x + C

2

Step VI : Replace u by y
x

( ) ( )1 ln2

2
y +1 + tan−1 y x = −ln x + C #
x

© Amirah 2022 251

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

Solution : Method II

Step I : write as a subject


2 − 2
= 2 + 2 + 2
Step II : Substitute = + and = ( )



( )2 − 2
+ = 2 + 2 ( ) + ( )2

2 2 − 2
+ = 2 + 2 2 + 2 2

Step III : factorize and simplify

2( 2 − 1)
+ = 2(1 + 2 + 2)

2 − 1
+ = 2 + 2 + 1

( + 1)( − 1)
+ = ( + 1)( + 1)

− 1
+ = + 1

Step IV : separating variables x and v

− 1
= + 1 −

− 1 − ( + 1)
=
+ 1

− 1 − 2 −
=
+ 1

−1 − 2
= + 1

−(1 + 2)
= + 1

+ 1 −1
1 + 2 =

© Amirah 2022 252

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

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

+ 1 −1
∫ 1 + 2 = ∫

1 1
∫ (1 + 2 + 1 + 2) = − ∫

1 1
∫ 1 + 2 + ∫ 1 + 2 = − ∫

1 2 1 1
2 ∫ 1 + 2 + ∫ 1 + 2 = − ∫

1 |1 + 2| + −1( ) = − +
2

Step VI : Replace v by ( )

1 |1 + ( )2| + −1 ( ) = − + #
2

Example 3 ( )ans : tan y x = ln x + C
Solve x dy =  y + x cos 2 y  dx
as a subject.
 x
Solution : Method I

Step I : write dy as a subject Homogeneous equation : write
dx

x dy =  y + x cos 2 y  dx
 x

 y + x cos2 y 
dy =  x

dx x

( ) ( )Step II : write dy into the form of y u= y and name as equation 
dx x x
, replace

( )dy
=y + x cos2 y
x

dx x x

( ) ( ) ( )dy = y
x
dx
y + cos2 y x → to write into the form of
x

dy = u + cos2 u ----------------  Where
dx and

Step IV : substitute  into  and separating variables x and u Differentiate (using product rule)
 =  : x du + u = u + cos2 u ----------------- 

dx

© Amirah 2022 253

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

x du = cos2 u
dx

1 du = 1 dx
cos2 u x

sec2 u du = 1 dx
x

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

 sec2 u du = 1 dx
x
tanu = ln x + C

Step VI : Replace u by y
x

( )tan y = ln x + C #
x

Solution : Method II

Step I : write as a subject



= ( + 2 ( ))

= + 2 ( )


Step II : Substitute = + and = ( )


+ = ( ) + 2( )


+ 2( )
+ =


Step III : factorize and simplify

( + 2( ))
+ =


+ = + 2 ( )


Step IV : separating variables x and v

= + 2( ) −


= 2( )


© Amirah 2022 254

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

11
2( ) =

2( ) = 1


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

∫ 2( ) = ∫ 1


( ) = +

Step VI : Replace v by ( )
( ) = + #



Example 4 – with initial condition

( )Solve x 3 + 6xy 2 y' = 8y 3 + 2x 2y , when x = 1, y = 2.

( ) ( )ans : ln
2yx 3 + y = ln x + ln18
x

Solution : Method I

Step I : write dy as a subject Homogeneous equation : write as a subject.
dx

( )x3 + 6xy 2 y' = 8y 3 + 2x 2y

( )x3 + 6xy 2 dy = 8y 3 + 2x 2y
dx

dy = 8y 3 + 2x 2y (homogeneous equation with degree 3)
dx x 3 + 6xy 2

( ) ( )Step II : write dy into the form of y u= y and name as equation 
dx x x
, replace

(Note : if the expression has the same degree, dividing each expression by x deg ree )

dy = 8y 3 + 2x2y ( )(dividing each term by x3) → to write into the form ofy
dx x3 x2x x

x3 + 6xy 2
x3 xx 2

( ) ( )dy8y 3 + 2 y
( )dx x
= x

1+ 6 y 2
x

© Amirah 2022 255

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

dy = 8u3 + 2u ----------------  Where
dx 1+ 6u 2 and

Step IV : substitute  into  and separating variables x and u Differentiate (using product rule)
----------------- 
=: x du + u = 8u3 + 2u
dx 1 + 6u2

x du = 8u3 + 2u −u
dx 1 + 6u2

( )x du 8u3 + 6u 2
dx
= 2u − u 1 +
1 + 6u2

x du = 8u3 + 2u − u − 6u3
dx 1 + 6u2

x du = 2u3 + u
dx 6u2 + 1

6u 2 + 1 du = 1 dx
2u3 + u x

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

 6u2 + 1 du = 1 dx
2u3 + u x

Therefore, the general solution is,
ln 2u3 + u = ln x + C

Step VI : Replace u by y
x

( ) ( )ln2y3+ y = ln x + C ------------------ 
x x

Step VII : Applying the given initial condition to find the value of C.
By substituting x = 1, y = 2 into ,

ln 2(2)3 + (2) = ln1 + C

ln18 = 0 + C

C = ln18 (into )

Thus, the particular solution is,

( ) ( )ln
2yx 3 + y = ln x + ln18 #
x

© Amirah 2022 256

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

Solution : Method II

Step I : write as a subject


( 3 + 6 2) ′ = 8 3 + 2 2

( 3 + 6 2) = 8 3 + 2 2


8 3 + 2 2
= 3 + 6 2

Step II : Substitute = + and = ( )


8( )3 + 2 2( )
+ = 3 + 6 ( )2

8 3 3 + 2 3
+ = 3 + 6 2 3

Step III : factorize and simplify

3(8 3 + 2 )
+ = 3(1 + 6 2)

8 3 + 2
+ = 1 + 6 2

Step IV : separating variables x and v

8 3 + 2
= 1 + 6 2 −

8 3 + 2 − (1 + 6 2)
=
1 + 6 2

8 3 + 2 − − 6 3
=
1 + 6 2

2 3 +
= 1 + 6 2

6 2 + 1 1
2 3 + =

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

6 2 + 1 1
∫ 2 3 + = ∫

|2 3 + | = +

© Amirah 2022 257

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

Step VI : Replace v by ( )
|2 ( )3 + ( )| = + #

Step VII : Initial condition = 1, = 2
ln |2 (2)3 + (2)| = (1) +

11

|16 + 2| = 0 +
= (18)

Particular solution :
|2 ( )3 + ( )| = + (18) #

© Amirah 2022 258

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

Example 5 – with initial condition

Solve (3x − y )dx + (x + y )dy = 0 , when x = 3 , y = 0.

( )ans : 1 ln y 2 + 3 + 1 tan−1 y  = ln 3 − ln x
2 x 3 3x

Solution : Method I

Step I : write dy as a subject Homogeneous equation : write as a subject.
dx

(3x − y )dx + (x + y )dy = 0
(x + y )dy = − (3x − y )dx
(y + x)dy = (y − 3x)dx

dy = y − 3x (homogeneous equation with degree 1)
dx y + x

( ) ( )Step II : write dy into the form ofy u= y and name as equation 
dx x x
, replace

(Note : if the expression has the same degree, dividing each expression by x deg ree )

dy = y − 3x ( )(dividing each term by x1) → to write into the form of y
dx x x x

y + x
x x

=( )dyy −3
( )dx x

y +1
x

dy = u − 3 ----------------  Where
dx u + 1 and

Step IV : substitute  into  and separating variables x and u Differentiate (using product rule)
 =  : x du + u = u − 3 ----------------- 

dx u + 1

x du = u − 3 − u
dx u + 1

x du = u − 3 − u(u + 1)

dx u + 1

x du = u − 3 − u2 − u
dx u + 1

x du = − u2 − 3
dx u + 1

© Amirah 2022 259

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

( )x du = − u2 + 3
dx u + 1

u + 1 du = − 1 dx
u2 + 3 x

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

 u +1 du = − 1 dx
u2 + 3 x

  u
u2 + 3 du
+ 1 = − 1 dx
u2 + 3 du x

  ( ) 12u du + 1 = − 1 dx
u2 + u2 + x
2
3 2 du
3

1 ln u2 + 3 + 1 tan−1 u  = − ln x +C
2 3 3

Step VI : Replace u by y
x

Therefore, the general solution is,

( ) ( )1 lny2+ 3 + 1 tan−1 y  = − ln x +C
x 3 x
2
3 

( )1 lny 2 + 3 + 1 tan−1 y  = − ln x + C ------------------ 
x 3 3x
2

Step VII : Applying the given initial condition to find the value of C.
By substituting x = 3 , y = 0 into ,

1 ln 02 + 3 + 1 tan−1(0) = − ln 3 + C

23
1 ln 3 + 0 = − ln 3 + C
2

1

ln 3 2 = − ln 3 + C

ln 3 = − ln 3 + C

( )C = ln 3 + ln 3 = 2 ln 3 = ln 3 2 = ln 3 (into )

Thus, the particular solution is,

( )1 lny 2 + 3 + 1 tan−1 y  = ln 3 − ln x #
x 3 3x
2

© Amirah 2022 260

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

Solution : Method II

Step I : write as a subject


(3 − ) + ( + ) = 0
( + ) = −(3 − )
−(3 − )
= +
− 3
= +

Step II : Substitute = + and = ( )


( ) − 3
+ = ( ) +

− 3
+ = +

Step III : factorize and simplify

( − 3)
+ = ( + 1)

− 3
+ = + 1

Step IV : separating variables x and v

− 3
= + 1 −

− 3 − ( + 1)
=
+ 1

− 3 − 2 −
=
+ 1

−3 − 2
= + 1

−(3 + 2)
= 1 +

1 + −1
3 + 2 =

© Amirah 2022 261

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

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

+ 1 −1
∫ 2 + 3 = ∫

1 1
∫ ( 2 + 3 + 2 + 3) = − ∫

1 1
∫ 2 + 3 + ∫ 2 + 3 = − ∫

1 2 1 1
2 ∫ 2 + 3 + ∫ 2 + (√3)2 = − ∫

1 | 2 + 3| + 1 −1 ( ) = − + #
2 √3
√3

Step VI : Replace v by ( )

1 |( )2 + 3| + 1 −1 (( )) = − +
2 √3 √3

1 |( )2 + 3| + 1 −1 ( ) = − + #
2 √3
√3

Step VII : Initial condition = √3, = 0

1 |( 0 2 + 3| + 1 −1 ( 0 ) = − √3 +
2 √3 √3√3
√3 )

1 |3| + 1 −1(0) = − √3 +
2 √3

321 + 0 = − √3 +
√3 = − √3 +
= 2 √3

2

= (√3)
= (3)

Particular solution :

1 |( )2 + 3| + 1 −1 ( ) = − + (3) #
2 √3 √3

© Amirah 2022 262

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

More Examples from Previous Semester Papers

 June 2019/ MAT235/ Q4c/ 8 marks

Find the general solution of the homogeneous differential equation dy = y 2 + x 2 + xy .
dx x 2
∶ −1 ( ) = +
Solution :

2 + 2 + =
=
2

( )2 + 2 + ( ) Find (using product rule)

+ = 2 = =
= + ′ = 1
2 2 + 2 + 2
+ = ′ =
2

2( 2 + 1 + )
+ =
2

+ = 2 + 1 +


= 2 + 1 + −


= 2 + 1


11
2 + 1 =

Integrating both sides
11

∫ 2 + 1 = ∫
−1( ) = +
−1 ( ) = + #



© Amirah 2022 263

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

 Dec 2018/ MAT235/ Q4c/ 8 marks

Find the general solution of the homogeneous differential equation dy = xy .
dx x 2 + 4y 2
1
Solution : ∶ 8( )2 − ( ) = +

=
= 2 + 4 2
( ) Find (using product rule)
+ = 2 + 4( )2
2
+ = 2 + 4 2 2 = =
2 = + ′ = 1
+ = 2(1 + 4 2)
′ =


+ = 1 + 4 2


= 1 + 4 2 −

− (1 + 4 2)
= 1 + 4 2

− − 4 3
= 1 + 4 2

−4 3
= 1 + 4 2

1 + 4 2 1
−4 3 =

Integrating both sides

1 + 4 2 1 1 ( )
∫ −4 3 = ∫ 8( )2 − = + #


1 4 2 1 1
∫ (−4 3 + −4 3) = ∫ 8( 22)
− ( ) = +

∫ (− 1 −3 − 1 = ∫ 1 1 − ( ) = +
4 ) (8 22)

1 −2 2 − ( ) = + #
− 4 ∙ −2 − = + 8 2

1
8 2 − = +

1 − ( ) = + #
8( )2

© Amirah 2022 264

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

 June 2018/ MAT235/ Q4c/ 8 marks

Find the general solution of the homogeneous differential equation dy = xy − 4y 2 .
1 dx x 2
∶ 4( ) = +
Solution :

− 4 2 =
= 2 Find (using product rule)

( ) − 4( )2
+ =
2
= +
2 − 4 2 2 = =
+ = ′ = 1
2
′ =
2( − 4 2)
+ =
2

+ = − 4 2


= − 4 2 −


= −4 2


11
−4 2 =

Integrating both sides

11
∫ −4 2 = ∫

∫ − 1 −2 = ∫ 1
4

1 −1
− 4 ∙ −1 = +

1
4 = +

1
4( ) = + #

1
(4 ) = +


4 = + #

© Amirah 2022 265

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 :

2 2 − 2 2 =
= 4 Find (using product rule)

2( )2 − 2 2
+ = 4 ( )

2 2 2 − 2 2 = + = =
+ = 4 2 ′ = 1

2(2 2 − 2) ′ =
+ = 4 2

2 2 − 2
+ = 4

2 − 1
+ = 2

2 − 1
= 2 −

2 − 1 − (2 )
=
2

2 − 1 − 2 2
=
2 Integrating both sides

− 2 − 1 −2 1
= 2 ∫ 2 + 1 = ∫

2 1 2 1
− 2 − 1 = − ∫ 2 + 1 = ∫

2 1 − | 2 + 1| = +
−( 2 + 1) = − |( )2 + 1| = + #

−2 1
2 + 1 =

© Amirah 2022 266

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

 Mar 2017/ MAT235/ Q4a/ 7 marks

( )Solve the homogeneous differential equation 3x3 + y 3 dy = 3x2y .
dx

Solution : ∶ 1 − ( ) = +
( )3

(3 3 + 3) = 3 2

=

3 2 Find (using product rule)
= 3 3 + 3

3 2( ) = =
+ = 3 3 + ( )3 = + ′ = 1

3 3 ′ =
+ = 3 3 + 3 3

3 3
+ = 3(3 + 3)

3 Integrating both sides
+ = 3 + 3
3 + 3 1
∫ − 4 = ∫
3
= 3 + 3 − 3 3 1

3 − (3 + 3) ∫ (− 4 + − 4) = ∫

= 3 + 3 ∫ (−3 −4 − 1 = ∫ 1
)
3 − 3 − 4
= 3 + 3 −3
−3 ∙ −3 − = +

− 4 −3 − = +
= 3 + 3
1
3 + 3 1 3 − = +

− 4 = 1 − ( ) = + #
( )3


© Amirah 2022 267

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

 Oct 2016/ MAT235/ Q5a/ 9 marks

Solve the first order homogeneous differential equation xy dy = 2y 2 − x2 .
dx

Solution : ∶ 1 |( )2 − 1| = +
2


= 2 2 − 2

=
2 2 − 2
= Find (using product rule)


2( )2 − 2 = =
+ = ( ) = + ′ = 1

2 2 2 − 2 ′ =
+ = 2

2(2 2 − 1)
+ = 2

2 2 − 1
+ =

2 2 − 1
= −

2 2 − 1 − ( )
=


2 2 − 1 − 2
=


2 − 1
=

1
2 − 1 =

Integrating both sides

1 = 2 − 1
∫ 2 − 1 = ∫ ∶ ∫ 2 − 1
= 2
1 | 2 − 1| = +
2 ∫ 2 − 1 = ∫ ∙ 2
= 2
1 |( )2 − 1| = + # 11
2 = 2 ∫

= 1 | | +
2

= 1 | 2 − 1| +
2

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

 Sep 2015/ MAT235/ Q4a(ii)/ 8 marks

Solve the first order homogeneous differential equation dy = x2−y 2 .
dx xy

Solution : ∶ 1 |1 − 2 ( )2| = +
− 4

2 − 2 =
=

2 − ( )2 Find (using product rule)
+ = ( )

2 − 2 2 = =
+ = 2 = + ′ = 1

2(1 − 2) ′ =
+ = 2

1 − 2
+ =

1 − 2
= −

1 − 2 − ( )
=


1 − 2 − 2
=


1 − 2 2
=
1
1 − 2 2 =

Integrating both sides = 1 − 2 2
∶ ∫ 1 − 2 2
1 = −4
∫ 1 − 2 2 = ∫
∫ 1 − 2 2 = ∫ ∙ −4
− 1 |1 − 2 2| = + = −4
4
11
1 |1 − 2 ( )2| = + # = − 4 ∫
− 4


= − 1 | | +
4

= − 1 |1 − 2 2| +
4

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