Question

What is the solution of D.E (D^2 – 2D)y = e^x  sin⁡x when solved using the method of undetermined coefficients?

(a) \( y =   c_1 +  c_2 \,e^{2x} –   \frac{e^x (xsin x + cos⁡x)}{2}\)

(b) \(y =   c_1 +  c_2 \,e^{2x} –   \frac{e^x sin x}{2}\)

(c) \(y =   c_1 +  c_2 \,e^{2x} –   \frac{e^x  cos x}{2}\)

(d) \(y =   c_1 +  c_2 \,e^{2x} –   \frac{e^x  (sin x + x cos⁡x)}{4}\)

This question was addressed to me in exam.

Question is from Method of Undetermined Coefficients topic in section Linear Differential Equations – Second and Higher Order of Engineering Mathematics

Answer 1

The correct choice is (b) \(y =   c_1 +  c_2 \,e^{2x} –   \frac{e^x sin x}{2}\)

The best explanation: A.E is m^2 – 2m = 0 or m(m – 2) = 0 –> m = 0,2

yc = c1 + c2 e^2x and ∅(x) =  e^x  sin⁡x. we assume PI in the form

yp = e^x (a cos x + b sin x)….(1) since 1±i are not roots of the A.E.

we have to find a, b such that yp” – 2yp‘ = e^x  sin⁡x…..(2)

from (1) yp‘ = e^x (- a sin x + b cos x) +  e^x (a cos x + b sin x)

yp” = e^x (- a sin x + b cos x) +  e^x (a cos x + b sin x) +  e^x (- a cos x –  b sin x) +  e^x (- a sin x + b cos x) = 2e^x (- a sin⁡x + b cos⁡x)  hence (2) becomes

2e^x (- a sin⁡x + b cos⁡x) – 2e^x (- a sin⁡x + b cos⁡x)

 – 2e^x (a cos⁡x + b sin⁡x) = e^x  sin⁡x i.e –  2ae^x  cos⁡x – 2be^x  sin⁡x =   e^x  sin⁡x

–> –  2a = 0, – 2b = 1 –> a = 0, b =  – 1/2 hence (1) becomes \(y_p = \frac{-e^x sin x}{2}\)

y = yc + yp = c1 + c2 e^2x – \(\frac{e^x sin x}{2}\) .