Question

Solution of the D.E y’’ – 4y’ + 4y = e^x when solved using method of undetermined coefficients is _____

(a) y = (c1 + c2)e^2x + 2e^x – 1

(b) y = (c1 + c2 x)e^2x + 4e^x – 4

(c) y = (c1 + c2 x)e^2x + e^x

(d) y = (c1 + c2 x)e^x + 4e^x

This question was posed to me during a job interview.

This is a very interesting question from Method of Undetermined Coefficients in chapter Linear Differential Equations – Second and Higher Order of Engineering Mathematics

Answer 1

Correct answer is (c) y = (c1 + c2 x)e^2x + e^x

For explanation: We have (D^2 – 4D + 4)y = e^x

A.E is m^2 – 4m + 4 = 0 –> (m – 2)^2 = 0 –> m = 2,2

thus yc =   (c1 + c2 x) e^2x, ∅(x) =  e^x and 1 is not a root of the A.E

we assume P.I in the form yp =   ae^x…(1) to find a such that yp’’ – 4yp’ + 4yp = e^x….(2)

yp’ = ae^x and yp’ = ae^x now (2) becomes ae^x – 4ae^x + 4ae^x = e^x –> a = 1

substituting the value of a in (1) we get yp = e^x

 thus the solution is y = yc + yp –> y = (c1 + c2 x) e^2x +  e^x.

Answer 2

The solution of the D.E y’’ – 4y’ + 4y = e^x when solved using the method of undetermined coefficients can be given by the Fourier differential method.