Question

Solve by the method of undetermined coefficients y" + y′ – 4y = x + cos2x.

Answer 1

We have 

(D² – D – 4)y = x + cos2x 

A.E. is 

m² – m – 4 = 0

We assume for P.I. in the form

yp = a + bx + c cos 2x + d sin 2x ...(1)

Since 0, ± i are not roots of the A.E. 

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

y"p – y′ p – 4yp = x + cos 2x ...(2)

From Eqn. (1)

y′ p = b – 2c sin 2x + 2d cos 2x

y"p = –4c cos 2x – 4d sin 2x

Now Eqn. (2) becomes,

– 4c cos2x – 4dsin2x – (b – 2csin2x + 2dcos2x) – [4 (a + bx + c cos2x + d sin2x)]

= x + cos2x

Comparing the coefficients, we have

– 4a – b = 0, – 4b = 1 and

– 8c – 2d = 1, 2c – 8d = 0

Solving these we get,