The P.I of (D^2+25)y=sin5x is
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The equation can be written as
(D2 – 5D + 6)y = e3x
(D – 3) (D – 2)y = e3x
C.F. = c1 e3x + c2e2x
And P.I. =1/ (D-3). 1/ (D-2) e3x
= 1/ (D-3) e2x ∫ e3x e-2xdx
=1/ (D-3) e2x ex
= e3x ∫ e3x e-3x dx = x.e3x
y = c1e3x +c2e2x +xe3x
P.I. can be found by resolving
1/f (D) = 1/ (D-3). 1/ (D-2)
Now using partial fractions,
1/f (D) = 1/ (D-3). 1/ (D-2)
= 1/ (D-3) – 1/ (D-2)
Hence, the required P.I. is [1(D-3) – 1/ (D-2)] e3x
= 1/ (D-3) e3x – 1/ (D-2) e3x
= e3x ∫ e3x e-3x dx – e2x ∫ e3x e-2x dx
= xe3x – e3x
Second term can be neglected as it is included in the first term of the C.F.
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