10 Differential Equation (Update)
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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
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• 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
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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
−=
+
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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
Replace x byReplace x by λλx and y byx and y by λλyy
( ) 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
Expand each termExpand each term
) ) x yλ λ
3 3 x / y 3 3
3 2
x e y
x y
λ λλ − λ=λ
Example 2 (cont.)Example 2 (cont.)
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
−=
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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
Replace x byReplace x by λλx and y byx and y by λλyy
3
2
x 3 y f ( x , y )
x y
−=
( ) ( ) 3
x 3 y f ( x , y )
λ − λλ λ =
Hasfazilah Ahmat
Jan 2010
Expand each termExpand each term
( ) ( ) x yλ λ
3 3
3 2 x 3 y
x yλ − λ=
λ
Example 3 (cont.)Example 3 (cont.)
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
−=
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• 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
−=
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Example 4 (cont.)Example 4 (cont.)
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− = +∫
Example 4 (cont.)Example 4 (cont.)
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= −
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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= −
Example 5 (cont.)Example 5 (cont.)
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 −
= +
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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
ReplaceReplace andand into theinto the
given problemgiven problem
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
+=
Example 6 (cont.)Example 6 (cont.)
Solve the initial value problemSolve the initial value problem 2 2 dy xy y 2 x
dx= + ( 1 ) 3=
o ut ono ut on 2 2 2
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= +
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Example 6 (cont.)Example 6 (cont.)
Solve the initial value problemSolve the initial value problem
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
Solve the initial value problemSolve the initial value problem
2 2( x 3 y )dx 2 xydy 0 , y( 1 ) 3− + = − = −
o ut ono ut on
ReplaceReplace andand into theinto the
given problemgiven problem
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
−=
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Example 7 (cont.)Example 7 (cont.)
Solve the initial value problemSolve the initial value problem 2 2( x 3 y )dx 2 xydy 0 , y( 1 ) 3− + = − = −
o ut ono ut on 2 2 2
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=∫ ∫
Example 7 (cont.)Example 7 (cont.)
Solve the initial value problemSolve the initial value problem
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
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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
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• 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− −= ∫
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Example 1 (cont.)Example 1 (cont.)
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=
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Example 2 (cont.)Example 2 (cont.)
Solve the differential equationSolve the differential equation
cos x dy
y cot x 4e dx + =
Rewrite the standard formRewrite the standard form
( ) 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
Determine integrating factorDetermine integrating factor
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−=
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Example 3 (cont.)Example 3 (cont.)
Solve the initial value problemSolve the initial value problem 2 2 dy x 6 x y , y( 0 ) 2
dx= + = −
Rewrite the standard formRewrite the standard form
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
− −= − +
Example 3 (cont.)Example 3 (cont.)
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
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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
Determine integrating factorDetermine integrating factor
Rewrite the standard formRewrite the standard form
( ) 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
= +∫
Example 4 (cont.)Example 4 (cont.)
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
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Example 4 (cont.)Example 4 (cont.)
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