10 Differential Equation (Update)

7/21/2019 10 Differential Equation (Update)

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Mat235 4/14/20

1 Differential Equations Differential Equations

Objectives:bjectives:

Understand some basic concepts andnderstand some basic concepts and

definitions concerning differential equationsefinitions concerning differential equations

Hasfazilah Ahmat

Jan 2010

definitions concerning differential equationsefinitions concerning differential equations

Identify homogeneous and linear equationsdentify homogeneous and linear equations

Solve homogeneous and linear equationsolve homogeneous and linear equations

Types of 1Types of 1stst DEDE

Hasfazilah Ahmat

Jan 2010



Mat235 4/14/20

• The differential equations

is homo eneous if the function

Homogeneous equationsHomogeneous equations

( ) dy

f x , y dx =

( f x , y

homogeneous, that is

( ) ( ) f x , y f x , y=λ λ

Hasfazilah Ahmat

Jan 2010

for any number λ

• Steps to check homogeneous equationshomogeneous equationhomogeneous equation

LetReplace x by

λx and y by λyExpand each

term( ) dy

f x , y dx

=

Hasfazilah Ahmat

Jan 2010

Factorize λ Cancel out λShow that

( ) ( ) f x , y f x , y=λ λ



Mat235 4/14/20

Example 1Example 1

Show that the differential equation is aShow that the differential equation is ahomogeneous equationhomogeneous equation 2

2 2

dy x xy

dx 3 x

−=

+

SolutionSolution

LetLet

Replace x byReplace x by λλx and y byx and y by λλyy

( ) 2

2 2

x xy f x , y

y 3 x

−=

+

( ) ( ) ( ) 2

x x y f ( x , y )

λ − λ λλ λ =

Hasfazilah Ahmat

Jan 2010

Expand each termExpand each term

( ) ( ) y 3 xλ + λ

2 2 2

2 2 2 2

x xy

y 3 x

λ − λ=λ + λ

Example 1 (cont.)Example 1 (cont.)

Show that the differential equation is aShow that the differential equation is a

homogeneous equationhomogeneous equation 2

2 2

dy x xy

dx 3 x

−=

+

FactorizeFactorize λλ22

Cancel outCancel out λλ22 ( ) 2

2 2

xy−=

( )( )

2 2

2 2 2

x xy

y 3 x

λ −=

λ +

Hasfazilah Ahmat

Jan 2010

ShownShown

Therefore,Therefore, is a homogeneousis a homogeneous

equationequation

f ( x , y )= 2

2 2

dy x xy

dx y 3 x

−=

+



Mat235 4/14/20

Example 2Example 2

Show that the differential equation is aShow that the differential equation is ahomogeneous equationhomogeneous equation 3 x / y 3

2

dy x e y

dx x

−=

SolutionSolution

LetLet


( ) 3 x / y 3

2

x e y f x , y

x y

−=

( ) ( ) 3 3 x / y

x e y f ( x , y )

λ λλ − λλ λ =

Hasfazilah Ahmat

Jan 2010


) ) x yλ λ

3 3 x / y 3 3

3 2

x e y

x y

λ λλ − λ=λ


Show that the differential equation is aShow that the differential equation is a

homogeneous equationhomogeneous equation 3 x / y 3

2

dy x e y

dx x y

−=

FactorizeFactorize λλ33

Cancel outCancel out λλ33

( )( )

3 3 x / y 3

3 2

x e y

x y

λ λλ −=

λ

3 x / y 3

2

x e y−=

Hasfazilah Ahmat

Jan 2010

ShownShown

Therefore,Therefore, is a homogeneousis a homogeneous

equationequation

f ( x , y )=

3 x / y 3

2

dy x e y

dx x y

−=



Mat235 4/14/20

Example 3Example 3

Verify that the equationVerify that the equation is not ais not ahomogeneous equation.homogeneous equation.

3

2

dy x 3 y

dx x y

−=

SolutionSolution

LetLet


3

2

x 3 y f ( x , y )

x y

−=

( ) ( ) 3

x 3 y f ( x , y )

λ − λλ λ =

Hasfazilah Ahmat

Jan 2010


( ) ( ) x yλ λ

3 3

3 2 x 3 y

x yλ − λ=

λ


Verify that the equationVerify that the equation is not ais not a

homogeneous equation.homogeneous equation.

3

2

dy x 3 y

dx x y

−=

FactorizeFactorize λλ

Cancel outCancel out λλ

( )( )

3

3 2

x 3 y

x y

λ −=

λ

( ) 3

2 2

3 y

x y

−=

λ

Hasfazilah Ahmat

Jan 2010

ShownShown

Therefore,Therefore, is not a homogeneousis not a homogeneous

equationequation

f ( x , y )≠

3

2

dy x 3 y

dx x y

−=



Mat235 4/14/20

• Steps to solve homogeneous equations

homogeneous equationhomogeneous equation

Let Differentiate both sideswrt x byusing productrule

Replace

and

Into the given

Applyseparableequation

y vx=

dy dvv x= +

y vx=

dy dvv x

dx dx= +

Hasfazilah Ahmat

Jan 2010

Example 4Example 4

Solve the homogeneous equation.Solve the homogeneous equation. 2 2 dy x 2 xy y

dx+ =

o ut ono ut on

ReplaceReplace andand into the giveninto the given

problemproblem

2 2 dy y x

dx 2 xy

−=

y vx= dy dv

v x dx dx

= +

Hasfazilah Ahmat

Jan 2010

( )( )

2 2vx x dvv x

dx 2 x vx

−+ =

2 2 2

2

v x x

2vx

−=



Mat235 4/14/20


Solve the homogeneous equation.Solve the homogeneous equation.

2 2 dy

x 2 xy y dx+ =

o ut ono ut on 2 2 2

2

v x x

2vx

−=

2v 1

2v

−=

2

2v 1 dv dx

x1 v− =

+∫ ∫

2u 1 v= +Let

Hasfazilah Ahmat

Jan 2010

dv v 1 2v x

dx 2v

− −=

21 v

2v

− −=

du 2vdv=

1 du ln x C

u− = +∫


Solve the homogeneous equation.Solve the homogeneous equation. 2 2 dy x 2 xy y

dx+ =

o ut ono ut on

1 du ln x C

u− = +∫

ln u ln x C − = +

2− =

2 C 1 v

x+ =

2

2

y C 1

x x= −

Hasfazilah Ahmat

Jan 2010

2ln 1 v ln x C + = − +

1 2ln 1 v ln x C −

+ = +

2 2 y Cx x= −



Mat235 4/14/20

Example 5Example 5

Solve the homogeneous equationSolve the homogeneous equation

y x dy y xe

dx x

−

=

o ut ono ut on

ReplaceReplace andand into theinto the

given problemgiven problem

vx= dy dv

v x dx dx

= +

vx x dv vx xe

v x −

+ =

Hasfazilah Ahmat

Jan 2010

x x

vv e= −

v dv x e

dx= −


Solve the homogeneous equationSolve the homogeneous equation y x dy y xe

dx x

−=

o ut ono ut on

v dv x e

dx= −

v

1 1 dv dx

xe− =

v

1 1 dv dx

xe− =∫ ∫

v 1−

Hasfazilah Ahmat

Jan 2010

x− =

ve ln x C

− = + y

xe ln x C −

= +



Mat235 4/14/20

Example 6Example 6

Solve the initial value problemSolve the initial value problem

2 2 dy

xy y 2 x dx= +

( 1 ) 3=

o ut ono ut on



vx= dy dv

v x dx dx

= +

2 2 dy y 2 x

dx xy

+=

Hasfazilah Ahmat

Jan 2010

( )

vx 2 x dvv x

dx x vx

++ =

2 2 2

2

v x 2 x

vx

+=


Solve the initial value problemSolve the initial value problem 2 2 dy xy y 2 x

dx= + ( 1 ) 3=


2

v x 2 x

vx

+=

2v 2

v

+=

v 1 dv dx

2 x=∫ ∫

2v

ln x C 4

= +

Hasfazilah Ahmat

Jan 2010

2 2 dv v 2 v x

dx v

+ −=

2

v=

v n x= + 2 y

4 ln x C x

⎛ ⎞= +⎜ ⎟

⎝ ⎠ 2 2 2

4 x ln x Cx= +



Mat235 4/14/20



2 2 dy

xy y 2 x dx= +

( 1 ) 3=

o ut ono ut on

2 2 2 y 4 x ln x Cx= +

This is the generalThis is the general

solution for the equationsolution for the equation

For particular solution, x= 1, y = 3For particular solution, x= 1, y = 3

Hasfazilah Ahmat

Jan 2010

3 4 1 ln 1 C 1= + C 9=

2 2 2 y 4 x ln x 9 x= +Therefore, the final solution isTherefore, the final solution is

Example 7Example 7


2 2( x 3 y )dx 2 xydy 0 , y( 1 ) 3− + = − = −

o ut ono ut on



vx= dy dv

v x dx dx

= +

2 2 dy 3 y x

dx 2 xy

−=

2 2−

Hasfazilah Ahmat

Jan 2010

( )v x

dx 2 x vx+ =

2 2 2

2

3v x x

2vx

−=



Mat235 4/14/20


Solve the initial value problemSolve the initial value problem 2 2( x 3 y )dx 2 xydy 0 , y( 1 ) 3− + = − = −


2

3v x x

2vx

−=

2 3v 1

2v

−=

2

2v 1 dv dx

xv 1=

−∫ ∫

2u v 1= −Let

Hasfazilah Ahmat

Jan 2010

2 2 dv 3v 1 2v

x dx 2v

− −=

2v 1

2v

−=

du 2vdv=

1 1 du dx

u x=∫ ∫



2 2( x 3 y )dx 2 xydy 0 , y( 1 ) 3− + = − = −

o ut ono ut on

1 1 du dx

u x=∫ ∫

ln u ln x C = + 2

2 2 3 y x Cx− =

The general solution forThe general solution for

the equation isthe equation is

For particular solution,For particular solution,

x=x= --1, y =1, y = --33

Hasfazilah Ahmat

Jan 2010

− 2v 1 Cx− = 2

y1 Cx

x

⎛ ⎞− =⎜ ⎟

⎝ ⎠

( ) ( ) ( ) 3 C 1 1− = − + −C 8= −

2 3 2 y 8 x x= − +

Therefore, the final solution isTherefore, the final solution is



Mat235 4/14/20

Types of 1Types of 1stst

DEDE

Hasfazilah Ahmat

Jan 2010

• A first order differential equations is said

to be linear if it can be written in the form

Linear equationsLinear equations

( ) ( ) dy

p x y q x dx

+ =

Hasfazilah Ahmat

Jan 2010



Mat235 4/14/20

• Steps to solve linear equations

Linear equationLinear equation

Rewrite the givenequation in standardform of linear equation Identify p(x)

Determine integratingfactor

( ) ( ) dy

p x y q x dx

+ = ( ) ( ) p x dx

I x e= ∫

Hasfazilah Ahmat

Jan 2010

ewr te t e stan arform as

mp y t e equat on,

( ) ( ) ( ) I x y I x q x dx= ∫ ( ) ( ) ( )1

y I x q x dx I x= ∫

Example 1Example 1

Solve the differential equationSolve the differential equation x 3 y 6 y 18e′ − =

divide by 3divide by 3

Identify p(x) =Identify p(x) = --22

Determine integrating factorDetermine integrating factor

x dy 2 y 6e

dx− =

2d x 2 x I( x ) e e− −∫= =

Hasfazilah Ahmat

Jan 2010

Rewrite the standard formRewrite the standard form

( ) ( ) ( ) I x y I x q x dx= ∫( ) 2 x 2 x x ye e 6e dx− −= ∫



Mat235 4/14/20


Solve the differential equationSolve the differential equation

x

3 y 6 y 18e′ − =

( ) 2 x 2 x x ye e 6e dx− −= ∫ 2 x x ye 6e dx− −= ∫ 2 x x− −

Hasfazilah Ahmat

Jan 2010

−

x

2 x6e c y

e

−

−− += x 2 x y 6e ce= − +

Example 2Example 2

Solve the differential equationSolve the differential equation cos x dy

y cot x 4e dx

+ =

IdentifyIdentify

Determine integrating factorDetermine integrating factor cot x d x

I( x ) e∫= cos x

dx sin xe= ∫

cos x dy cot x 4e

dx+ =

( ) p x cot x=

Hasfazilah Ahmat

Jan 2010

u s n x du cos x dx

==

1 du

ue= ∫ ln ue=lnsinxe= sin x=



Mat235 4/14/20


Solve the differential equationSolve the differential equation

cos x dy

y cot x 4e dx + =


( ) cos x y sin x sin x 4e dx= ∫ u cos x

du sin x dx

=

= −u y sin x 4e du= −∫

u

Hasfazilah Ahmat

Jan 2010

= − cos x y sin x 4e C = − + cos x 4e C

y sin x

− +=

cos x y 4 cos ec x e C cos ec x= − +

Example 3Example 3

Solve the initial value problemSolve the initial value problem 2 2 dy

x 6 x y , y( 0 ) 2 dx

= + = −

o ut ono ut on

IdentifyIdentify


2 2 dy6 x y x

dx− =

( ) 2 p x 6 x= −

26 x d x−∫

Hasfazilah Ahmat

Jan 2010

3 2 x I( x ) e−=



Mat235 4/14/20


Solve the initial value problemSolve the initial value problem 2 2 dy x 6 x y , y( 0 ) 2

dx= + = −


3

2

u 2 x

du 6 x dx

= −

= −

3 2 x I( x ) e−=

3 3 2 x 2 x 2 ye e x dx− −= ∫ 3

Hasfazilah Ahmat

Jan 2010

ye e du6

− = −

3 2 x u1 ye e c

6

− = − + 3 3 2 x 2 x1

ye e c6

− −= − +


Solve the initial value problemSolve the initial value problem 2 2 dy

x 6 x y , y( 0 ) 2 dx

= + = −

3 3 2 x 2 x1 ye e c

6

− −= − +

3

3

2 x

2 x

1e c

6 ye

−

−

− +=

3 2 x1 y ce

6 = − +

General solution

Hasfazilah Ahmat

Jan 2010

Given x = 0, y = -2

( ) 3

2 01 2 ce

6

−− = − + 11

c6

= − 3 2 x1 11 y e

6 6 = − −

Particular solution



Mat235 4/14/20

Example 4Example 4

Solve the initial value problemSolve the initial value problem dy tan x cos x , y( 0 ) 4

dx− = =

o ut ono ut on

IdentifyIdentify



( ) p x tan x= −

tan x d x I( x ) e

−∫= lncos xe= cos x=

Hasfazilah Ahmat

Jan 2010

y cos x cos x cos x dx= ∫ 2 y cos x cos x dx= ∫

( )1

1 cos 2 x dx 2

= +∫


Solve the initial value problemSolve the initial value problem dy

tan x cos x , y( 0 ) 4 dx

− = =

o ut ono ut on

( )1

1 cos 2 x dx 2

= +∫1 sin 2 x

y cos x x C 2 2

⎛ ⎞= + +⎜ ⎟

⎝ ⎠

Hasfazilah Ahmat

Jan 2010

y x sec x C sec x 2 4 cos x= + +

Given x = 0, y = -2

( ) ( ) ( )

( ) ( )

sin 2 01 1 4 0 sec 0 C sec 0

2 4 cos 0= + + C 4=

General solution



Mat235 4/14/20


Solve the initial value problemSolve the initial value problem dy tan x cos x , y( 0 ) 4

dx− = =

o ut ono ut on

Given x = 0, y = -2

( ) sin 2 01 1 4 0 sec 0 C sec 0= + + C 4=

1 1 sin 2 x y x sec x C sec x

2 4 cos x= + +

Hasfazilah Ahmat

Jan 2010

2 4 cos 0

1 1 sin 2 x y x sec x 4 sec x

2 4 cos x= + +

Particular solution

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10 Differential Equation (Update)