Question

Solve the Ordinary Differential Equation y’’ + 2y’ + 5y = e^-t sin(t) when y(0) = 0 and y’(0) = 1.(Without solving for the constants we get in the partial fractions).

(a) \(e^t [Acost+A1sint+Bcos(2t)+\frac{(B1)}{2} sin(2t)] \)

(b) \(e^{-t} [Acost+A1sint+Bcos(2t)+B1sin(2t)] \)

(c) \(e^{-t} [Acost+A1sint+Bcos(2t)+\frac{(B1)}{2} sin(2t)] \)

(d) \(e^t [Acost+A1sint+Bcos(2t)+(B1)sin(2t)] \)

I had been asked this question during an interview for a job.

Enquiry is from Solution of DE With Constant Coefficients using the Laplace Transform in portion Laplace Transform of Engineering Mathematics

Answer 1

Right option is (c) \(e^{-t} [Acost+A1sint+Bcos(2t)+\frac{(B1)}{2} sin(2t)] \)

The explanation is: \(L[y’’+2y’ +5y = e^{-t} sin(t)] \)

\(s^2 Y(s)-sy(0)-y'(0)+ 2sY(s) -2y(0) + 5Y(s) = \frac{1}{(s+1)^2+1} \)

\((s^2+2s+5)Y(s)= \frac{1}{(s+1)^2+1}+1 \)

\((s^2+2s+5)Y(s)= \frac{(s^2+2s+3)}{(s^2+2s+2)} \)

\( Y(s) = \frac{(s^2+2s+3)}{(s^2+2s+2)(s^2+2s+5)} \)

\( = \frac{(s+1)^2+2}{((s+1)^2+1)((s+1)^2+4)} \)

\( y(t) = e^{-t} L^{-1} [\frac{(As+A1)}{(s^2+1)}+\frac{(Bs+B1)}{(s^2+4)}] \)

\( = e^{-t} [Acost+A1sint+Bcos(2t)+\frac{(B1)}{2} sin(2t)]\).