Particular integral (P.I.) of (D^2+ 1)y = x A) x/2 B) x2 C) x D) x2 - 1 full solve
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Answer:
We have to solve (D
2
−4D+1)y=x
2
The characteristic equation is p
2
−4p+1=0
⇒p=
2
4±
16−4
=
2
4±2
3
=2±
3
Thus Complementary function C.F.=Ae
(2+
3
)x
+Be
(2−
3
)x
Particular integral P.I.=
D
2
−4D+1
1
(x
2
)
=[1−(4D−D
2
)]
−1
(x
2
)
=[1+(4D−D
2
)+(4D−D
2
)
2
+...](x
2
)
=[1+4D+15D
2
+...](x
2
)
∴P.I.=x
2
+8x+30
Hence the general solution is y=C.F.+P.I.
y.=Ae
(2+
3
)x
+Be
(2−
3
)x
+(x
2
+8x+30)
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