L-6]

Method : When

Let be the differential equation then

f'o V.
patcula integral
fCO)
(rco]
Problems:
Solve (0+4)9
O+)y-sinx
Sol: Given
) VSin x.
fco) - 0+4

ci) foding complementary Solution Cyc)i

Auri lary equ i fcm) =o )

m= +2i
=)

C,eos 2T + C29in2x

(") finding paiculan inteqral s Olution (yp):

20 Sinx
Sint

sin 2D sinx

-i+4 (-i)
sìnx 2D sint

TSint 2 Ccosr)
3

2 Cosr.
y= G COS2X + C2 Sin 20 3
2.
Solve (D'-)y Sin 37 + COS x.

Sol: Given D.£ is (D'-)y =

XSin31 + COSX.
fco) = D 8(1)=

cæ) finding c i
Aurilary equaton fcm) = o

m- =0 m=

finding yp:
Sin 30 + CoSx

COS
+

COS X
+
Sin3x - 2D Sin3x
(D)

Sin3x 2D Sin 3x
--1
-2+)

a Singx 2 D sinn
-2
-10

CoS X
T Sin3x 2 (3 (OS)X)

3 CoS3x
y 4e4 Che
50
 3. Solve
aos.
DE is (o'-)9 =
Sol: Given
fco) - o' Q() =

m- -0
Auzilary equation fem) 0
G,e + e

(i) finding yp:

Yp = cosx (rcosx)

/" RP of e Cos ]
D

D'-I
ix
R.P of e
(D+)-i

R.P of e
D42i0-2
2

R.P of e
-
2

1
D'+2iD
RP of e

R.P of e
 R.P of e D-4o°+yi D
- Rp of e iD + -D+

ix
RP of e i(23) - 2

ix
RP of e a+ 2ix -)
2

(Cos* ++
isinx) Cx-) + +2ix]

-) cosz 2sinx
(-x) cost +7Sin.
2

I (-2) cos..
y 2

2
Solve (0-40 +)y Sio2x.
21
Sol: Given DE is
2T
8xe Sin 20.
fco)= D-yD +4
findng Yci
m - ymty 0
ilory
Auxiloy equ: (m-2)=o
m: 2, 2
=)

(G + x)e.
 fnding ypi
yp 82 e Sin 21
(o-2)>

2
(D+2 -2)

8e2

21
8e T.P of1
D2

Tmp P &s e (0+2i)

2

22 i21
8ef.P of e

8e2r 2i
1.P of e 2:0 +3.?
2i

() (*t
22
8e T.P of e

2 2)X
8e .S.P of e

2e
22
îP Of e

2e2r

y=