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We assume a particular solution of
1 2cos sin
ax ax
Ly k e bx k e bx as 1 2cos sin
ax ax
py y A e bx A e bx
We assume a particular solution of
0 1( ... )
ax n
nLy e b b x b x
as 0 1( ... )ax n
p ny y e A A x A x
We also use multiply by the least power ofx
rule in case any term of the Given RHS is a
part of the C.F.
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Example8 Find the general solution of
the d.e.
2 2 sinx
y y y e x Solution We first find the complementary
function.
Auxiliary Equation: 2 2 2 0m m Roots: m =
Hence theComplementary function is
1 2( cos sin ),
x
hy e c x c x c
1, c
2arbitrary constants
1 i
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Now we find the particular solution.
As sinx
e x is a part of the C.F., we take a
particular solution as
1 2( cos sin )
x xy x A e x A e x
1 2 2 1[( ) cos ( ) sin ]x xDy x A A e x A A e x
1 2( cos sin )
x xA e x A e x
2
2 1[2 cos 2 sin ]x xD y x A e x A e x 1 2 2 1[2( ) cos 2( ) sin ]
x xA A e x A A e x
2
-2
1
Adding, we get
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2( 2 2)D D y
2 12 cos 2 sin
x xA e x A e x
sin
x
e x gives
1 22 1, 2 0A A
1 2
1, 0
2A A or
Hence a particular solution is cos2
x
p
xy e x
Hence the general solution ish p
y y y
i.e. 1 2( cos sin )x
y e c x c x cos2
xxe x
c1, c2 arbitrary constants
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The method of Variationof Parameters
In this lecture we discuss a general method offinding a particular solution of a non-
homogeneous l.d.e. (whether it is constant
coefficient equation or not).
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Consider the second order l.d.e.
( ) ( ) ( )..(*)y P x y Q x y h x
We assume that we have already found the C.F.
as
1 1 2 2( ) ( )y c y x c y x
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The Method of Variation of parameters says:
Take a particular solution of (*) as
1 1 2 2( ) ( )y v y x v y x (**)
where v1(x), v2(x) are functions ofx to bechosen such that (**) is a solution of (*).
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1 1 2 2( ) ( )y v y x v y x (**)
Differentiating (**) w.r.t. x, we get
1 1 2 2 1 1 2 2( ) ( ) ( ) ( )y v y x v y x v y x v y x
We now choose v1, v2 such that
1 1 2 2( ) ( ) 0 .....(1)v y x v y x
Thus1 1 2 2
( ) ( )y v y x v y x
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Differentiatingy w.r.t. x, we get
1 1 2 2 1 1 2 2( ) ( ) ( ) ( )y v y x v y x v y x v y x
Substituting fory,y,y in (*), we get
1 1 2 2 1 1 2 2
1 1 2 2
1 1 2 2
( )
( ) ( )
v y v y v y v y
P x v y v y
Q x v y v y h x
The coefficients ofv1
, v2
are zero as
1 1 2 2( ) ( )y v y x v y x
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y1,y2 are solutions of the associated
homogeneous l.d.e.
( ) ( ) 0y P x y Q x y
Thus we get
1 1 2 2( ) ( ) ( ) ....(2)v y x v y x h x
Solving (1), (2) we get 1 2,v v
Integrating, we get 1 2,v v
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The two equations satisfied by 1 2,v v
are1 1 2 2( ) ( ) 0 ...........(1)v y x v y x
1 1 2 2( ) ( ) ( ) ....(2)v y x v y x h x
We note that the determinant of the coefficient
matrix is
1 2
1 2
( ) ( )
( ) ( )
y x y x
y x y x
1 2[ , ]( )W y y x 0
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asy1,y2 are LI solutions of the associated
homogeneous l.d.e. Using Cramers rule, we get
2
2
11 2
0 ( )
( ) ( )
[ , ]( )
y x
h x y x
v W y y x
2
1 2
( ) ( )
[ , ]( )
h x y x
W y y x
1
1
2
1 2
( ) 0
( ) ( )
[ , ]( )
y x
y x h xv
W y y x
1
1 2
( ) ( )[ , ]( )
h x y xW y y x
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Integrating, we get 21
1 2
( ) ( )
[ , ]( )
h x y xv dx
W y y x
12
1 2
( ) ( )
[ , ]( )
h x y xv dx
W y y x
And hence a particular solution is
1 1 2 2( ) ( )py y v y x v y x
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Example 1 Find the general solution of the d.e.
4 sec 2y y x Auxiliary Equation
24 0m
Roots: m = 2i
Hence the complementary function is
1 2cos 2 sin 2hy y c x c x
c1, c2 arbitrary constants
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Hence we take a particular solution as
1 2cos 2 sin 2y v x v x where v1(x), v2(x) are functions ofx to be
chosen such that the above is a solution of
the given d.e.
Differentiating w.r.t. x, we get
1 2( 2sin 2 ) (2 cos 2 )y v x v x
1 2cos 2 sin 2v x v x
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We now choose v1, v2 such that
1 2cos 2 sin 2 0 .....(1)v x v x
Differentiatingy w.r.t. x, we get
1 2( 2 sin 2 ) (2 cos 2 )y v x v x
1 2( 4 cos 2 ) ( 4 sin 2 )y v x v x
1 2( 2 sin 2 ) (2 cos 2 )v x v x
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Substituting fory,y,y in the given d.e.,
we get4 sec 2y y x
1 2( 4 cos 2 ) ( 4 sin 2 )v x v x
1 2( 2 sin 2 ) (2 cos 2 )v x v x
1 2
4( cos 2 sin 2 ) sec 2v x v x x
1 2( 2sin 2 ) (2 cos 2 ) sec 2v x v x x
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The two equations satisfied by 1 2,v v are
1 2cos 2 sin 2 0 .....(1)v x v x 1 2
( 2sin 2 ) (2 cos 2 ) sec 2 ...(2)v x v x x
Using Cramers rule, we get
We note that the determinant of the coefficientmatrix is
cos 2 sin 2
2sin 2 2 cos 2
x x
x x 2 0
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1
0 sin 2
sec 2 2 cos 2
2
x
x xv
1tan2
2x
2
cos 2 0
2sin 2 sec 2
2
x
x xv
12
Integrating, we get1
ln(sec 2 )4
x 2v 1
ln(cos 2 ),4
x1
v 1
2x
1 1
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Hence a particular solution is
1 2cos 2 sin 2y v x v x 1 1
[ln(cos 2 )]cos 2 sin 24 2
x x x x
Hence the general solution is
1 2. . cos 2 sin 2i e y c x c x
c1, c2 arbitrary constants
h py y y
1 1[ln(cos 2 )]cos 2 sin 2
4 2x x x x
1
1ln(cos 2 ),
4v x
2
1
2v x
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Example 2 Find the general solution of the d.e.
2 2( 1) (2 ) (2 ) ( 1)x x y x y x y x We first find the complementary function .
1
x
y y e
We assume a second LI solution as
As the sum of the coefficients (of the LHS)2
( 1) (2 ) (2 ) 0x x x x
is one solution of the C.F.
2 1y y vy where
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2
1
1 P dxv e dx
y
2
(2 )
( 1)
2
1x
dxx x
x e dxe
Now
2
2( 1)
xx x
21
( 1)x
x x
2 111x x
Hence 2(2 )( 1)
xdx
x xe
2 1
(1 )1
dxx xe
2
1 xxe
x
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2(2 )
( 1)
2
1x
dxx x
xv e dx
e
Thus
2 2
1 1 xx
xe dx
e x
2
1 1( )
xe dx
x x
1 x
ex
Hence2 1
y vy 1x
And hence 2 1yx
is a second LI solution
of the associated homogeneous d.e.
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Hence the C.F. is 1 21x
y c e cx
(c1, c2 arbitrary constants)So let a particular solution be 1 2
1xy v e v
x
Differentiating w.r.t. x, we get
1 2 1 22
1 1( )
x xy v e v v e v
x x
We now choose v1, v2 such that
1 2
10 ....(1)
xv e vx
Hence1 2 2
1( )
xy v e v
x
1
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Differentiatingy w.r.t. x, we get
1 2 1 23 22 1( ) ( )x xy v e v v e vx x
Substituting fory,y,y in the given d.e.,
1 2 1 23 2
2 1( 1)[ ( ) ( )]
x xx x v e v v e v
x x
2
1 2
1(2 )[ ] ( 1)
xx v e v x
x
2
1 2 2
1
(2 )[ ( )]x
x v e v x
1 2 2
1( )
xy v e vx
we get
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1 2 2
1 ( 1)( ) ....(2)
x xv e v
x x
The two equations satisfied by 1 2,v v are
1 2
10 ....(1)
xv e v
x
1 2 2
1 ( 1)( ) ....(2)
x xv e v
x x
We note that the determinant of the coefficientmatrix is
2
1 1xe
x x
2
1x xe
x
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Using Cramers rule, we get
2
1
2
1
0
( 1) 1
1( )
x
x
x
x xv xe
x
xe
2
2
0
1
1( )
x
x
x
e
xe
x
v xe
x
x
Integrating, we get 1v ,xe 2v
2
2
x
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Hence a particular solution is
1 2
1xy v e v
x
12
x
2
1 2,
2
x xv e v
Hence the general solution ish p
y y y
i.e. 1 21x
y c e c x 1 2
x
c1, c2 arbitrary constants
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Example 3 Find the general solution of the d.e.
2 ln .xy y y e x Auxiliary Equation
22 1 0m m
Roots: m = 1, 1 Hence the complementary function is
1 2( )
x x
hy y c e c x e
c1, c2 arbitrary constants
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Hence we take a particular solution as
1 2
x x
y v e v xe
where v1(x), v2(x) are functions ofx to be
chosen such that the above is a solution of the
given d.e.
Thus here 1 2,x x
y e y xe
Wronskian = W = x x
x x x
e xe
e e xe
2xe
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Noting that ( ) lnx
h x e x we get
21
1 2
( ) ( )
[ , ]( )
h x y xv dx
W y y x
121 2
( ) ( )
[ , ]( )
h x y xv dx
W y y x
2
lnx x
x
e x xedx
e
lnx x dx 2 2ln
2 4
x xx
2
lnx x
x
e x edx
e
ln x dx lnx x x
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Hence a particular solution is
1 2
x x
py y v e v xe
2 2( ln )
2 4
xx xx e
( ln ) xx x x xe
223
( ln )2 4
xx x x e
And the general solution is h py y y i.e.
1 2
x xy c e c x e
2
23( ln )
2 4
xxx x e
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Example 4 Find the general solution of the d.e.2
2 2 .x
x y x y y x e
Solution Consider the associated homogeneous
equation
zx e
Note that2
2,dy d y dyxy xy
dz dz dz
2
2 2 0x y x y y
We put
(**)
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Auxiliary equation2
3 2 0m m Roots m = 1, 2
Solution of Eqn (**) is2
1 2z zy c e c e
2
1 2y c x c x
c1, c2 arbitrary constants
i.e. The complementary function of the
given d.e. is
Hence the equation (**) becomes
2
( 3 2) 0y ( )
d
dz
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So we take a particular solution as2
1 2y v x v x Thus here
2
1 2,y x y x
Wronskian = W =2
1 2
x x
x
2
x
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Noting that ( )x
eh x
x
We get
2
1
1 2
( ) ( )
[ , ]( )
h x y xv dx
W y y x
1
2
1 2
( ) ( )
[ , ]( )
h x y xv dx
W y y x
xedx
x
2
xedx
x
x x
e edx
x x
i l l i i
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Hence a particular solution is
2
1 2py y v x v x
2x x
xe ex dx xe x dx
x x
And the general solution is h py y y
i.e.
2
1 2y c x c x
2( )
xx exe x x dx
x
2( )
xx e
xe x x dxx