Solution To fi nd the complementary function, we have to solve
2
(44)0DDy++=
The auxiliary equation is
22
2
12
44(2)02,2
()
x
c
mmmm
yccxe
−
++=+=⇒=−−
=+
To fi nd the particular integral,
2
2
22
22
2
1
coscos
(2)(22)
1
coscos
x
x
p
xx
e
yexx
DD
exex
D
−
−
−−
==
+−+
==−
The general solution is
22
12
()()()cos.
xx
cp
yxyxyyccxeex
−−
==+=+−
EXERCISE 3.8
Solve the following:
1.
22
(76)(1).
x
DDyex−+=+ [JNTU 2003]
Ans:
2
6
12
(41)
16
x
xx
e
ycecex
−−
=+−+
2.
2
(21).
x
DDyxe−+=
Ans:
12
()(sin2cos)
xx
yccxeexxx=+−+
3.
322
(3 ...
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