Question

What  is the inverse Laplace Transform of a function y(t) if after solving the Ordinary Differential Equation Y(s) comes out to be \(Y(s) = \frac{s^2-s+3}{(s+1)(s+2)(s+3)} \) ?

(a) \(\frac{1}{2} e^{-t}+\frac{9}{2} e^{-3t}-3e^{-2t} \)

(b) \(\frac{-1}{2} e^{-t}+\frac{9}{2} e^{-2t}-3e^{-3t} \)

(c) \(\frac{1}{2} e^{-t}-\frac{3}{2} e^{-2t}-3e^{-3t} \)

(d) \(\frac{-1}{2} e^{t}+\frac{9}{2} e^{2t}-3e^{3t} \)

This question was addressed to me in semester exam.

Question is from Solution of DE With Constant Coefficients using the Laplace Transform in division Laplace Transform of Engineering Mathematics

Answer 1

Right choice is (b) \(\frac{-1}{2} e^{-t}+\frac{9}{2} e^{-2t}-3e^{-3t} \)

The explanation: Taking inverse Laplace Transformation for

\(Y(s) = \frac{(s^2-s+3)}{(s+1)(s+2)(s+3)} \)

 Solving the partial fractions we get,

\(Y(s) = \frac{-1}{2}  \frac{1}{(s+1)}+\frac{9}{2}  \frac{1}{(s+2)}-3 \frac{1}{(s+3)} \)

Therefore, \(y(t) = \frac{-1}{2} e^{-t}+\frac{9}{2} e^{-2t}-3e^{-3t}. \)