Question

Solve by the method of undetermined coefficients y" + y′ – 2y = x + sin x.

Answer 1

We have 

(D² + D – 2) y = x + sin x 

A.E. is m² + m – 2 = 0 

i.e., (m – 1) (m + 2) = 0, 

m = 1, – 2 

C.F. = C₁e^x + C₂e^{– 2x} 

φ (x) = x + sin x and 0 is not root of the A.E

We assume for P.I. in the form

y_p = a + bx + c cos x + d sin x. ...(1)

We have to find a, b, c, and d such that

y"_p + y′_p – 2y_p = x + sin x ...(2)

From (1),

y′_p = b – c sin x + d cos x

y"_p = – c cos x – d sin x

Eqn. (2), becomes

– cos x – d sin x + b – c sin x + d cos x – 2 (a + bx + c cos x + d sin x) = x + sin x (– 2a + b) – 2bx + (– 3c – d ) sin x + (c – 3d) cos x = x + sin x

Comparing the coefficients, we get

– 2a + b = 0, – 2b = 1, – 3c – d = 1, c – 3d = 0

Solving, we get