Question

Solve by the method of undetermined coefficients (D² + 1)y = 4x – 2sinx.

Answer 1

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 

y_p = 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 + y_p = 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.