Question

Find the \(L^{-1} (\frac{1}{s(s+4)^\frac{1}{2}})\), give the answer in terms of error function.

(a) \(\frac{1}{2} erf⁡(2t)\)

(b) \(\frac{1}{2} erf⁡(\sqrt{t})\)

(c) \(\frac{1}{2} erf⁡(2\sqrt{t})\)

(d) \(\frac{1}{2} erf⁡(4\sqrt{t})\)

The question was asked in final exam.

My question comes from Convolution in division Laplace Transform of Engineering Mathematics

Answer 1

Correct choice is (c) \(\frac{1}{2} erf⁡(2\sqrt{t})\)

For explanation I would say: In the given question,

Let \(p_1(s) = \frac{1}{(s+4)^{\frac{1}{2}}}\) and \(p_2(s) = \frac{1}{s}\)

\(f_1 (t) = L^{-1} (\frac{1}{(s+4)^{\frac{1}{2}}}) = \frac{e^{-4t}×\sqrt{t}}{\sqrt\pi}\)

\(f_2 (t) = L^{-1} \frac{1}{s} = 1\)

By Convolution Theorem,

\(L^{-1} (p_1(s)×p_2(s)) = \int_{0}^{t}f_1(u) f_2(t-u) dt\)

\(L^{-1} (\frac{1}{s(s+4)^{\frac{1}{2}}}) = \int_{0}^{t}\frac{\sqrt{u}}{\sqrt\pi} e^{-4u} \,du\)

\(=\frac{1}{\sqrt\pi} \int_{0}^{t}\sqrt{u} e^{-4u} du\)

We know that error function is given by –

\(erf⁡(x)=\frac{2}{\sqrt\pi} \int_{0}^{x}e^{-z^2} dz\)

Applying, 4u=z^2 And setting the limits of u in terms of z, we get

\(L^{-1} (\frac{1}{s(s+4)^{\frac{1}{2}}}) = \frac{1}{2} erf⁡(2\sqrt{t})\)

Thus, the answer is \(\frac{1}{2} erf⁡(2\sqrt{t})\).