Question

Find the inverse laplace transform of \(\frac{1}{s(s-1)(s^2+1)}\).

(a) ^1⁄2 e^-t + ^1⁄2 Sin(-t) – ^1⁄2 Cos(-t)

(b) ^1⁄2 e^t + ^1⁄2 Sin(t) – ^1⁄2 Cos(t)

(c) ^1⁄2 e^t + ^1⁄2 Sin(t) + ^1⁄2 Cos(t)

(d) ^1⁄2 e^t – ^1⁄2 Sin(t) – ^1⁄2 Cos(t)

I have been asked this question in an online quiz.

My enquiry is from Laplace Transform by Properties topic in chapter Laplace Transform of Engineering Mathematics

Answer 1

Correct choice is (b) ^1⁄2 e^t + ^1⁄2 Sin(t) – ^1⁄2 Cos(t)

For explanation: We know that,

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

Let, G(s)=\(\frac{1}{(s-1)(s^2+1)}=\frac{1}{2(s^2-1)}-\frac{s+1}{2(s^2+1)}=\frac{1}{2*(s-1)}-\frac{s}{2(s^2+1)}-\frac{1}{2(s^2+1)}\)

Now, g(t)=\(\frac{1}{2}e^t-\frac{1}{2}cos(t)-\frac{1}{2}cos(t)\)

Now, Y(s)=\(\frac{1}{2}G(s)=>y(t)=\int_0^t g(t)dt=\frac{1}{2}e^t+\frac{1}{2}sin(t)-\frac{1}{2}cos(t)\)