A.E. is
m2 + 1 = 0 ⇒ m = ± i
∴ C.F. = C1cos x + C2sin x
φ (x) = 4x – 2 sin x. We assume for P.I in the form
yp = a + bx + x (c cos x + d sin x) ...(1)
Since 0 is not the root of the A.E. and ± i is the root of A.E.
We have to find a and b such that
y"p + yp = 4x – 2sin x ...(2)
From Eqn. (1) y′p = b + x (– c sin x + d cos x) + (c cos x + d sin x)
y"p = x (– c cos x – d sin x) + (– c sin x + d cos x) + (– c sin x + d cos x)
Eqn. (2), becomes
x (– c cos x – d sin x) + (– c sin x + d cos x) + (– c sin x + d cos x) a + bx + x (cos x + d sin x) = 4x – 2 sin x
i.e.,a + bx + (– 2c sin x + 2d cos x) = 4x – 2 sin x
Comparing the coefficients, we get,
a = 0, b = 4, – 2c = – 2, 2d = 0
i.e., a = 0, b = 4, c = 1, d = 0
Therefore the required P.I. (using Eqn. (1)
From Eqn. (1) ⇒ P.I. = 4x + x cos x
∴ y = C.F. + P.I.
= C1 cos x + C2 sin x + 4x + x cos x.
