Question

L-¹(1/s+a)(s+b) by convolution theorem​

Answer 1

CALCULATION

We can observe that

$\longmapsto \rm \: \dfrac{1}{(s + a)(s + b)} = \dfrac{1}{s + a} \times \dfrac{1}{s + b}$

Also,

We know that

$\longmapsto \rm \: {L}^{ - 1} \bigg(\dfrac{1}{s + a} \bigg) \: = \: {e}^{ - at}$

and

$\longmapsto \rm \: {L}^{ - 1} \bigg( \dfrac{1}{s + b} \bigg) \: = \: {e}^{ - bt}$

In other words, we are able to factorize the given expression  into a product of two expressions whose Inverse Laplace Transforms are already known to us. Whenever we see this situation of the expression in question, the solution is a straight forward application of the Convolution Theorem.

So,

By Convolution Theorem, we have

$\rm \: {L}^{ - 1} \bigg( F(s) \: \times \: G(s)\bigg) \: = \: \int_0^t \: f(u) \: \times \: g(t - u) \: du$

where f(t) and g(t) are the Inverse Laplace Transforms of F(s) and G(s), respectively. In other words,

$\rm \: {L}^{ - 1} \bigg(\dfrac{1}{(s + a)(s + b)} \bigg) = \rm \: {L}^{ - 1} \bigg( \dfrac{1}{s + a} \times \dfrac{1}{s + b} \bigg)$

$\longmapsto \rm \: = \int_0^t \: {e}^{ - au} {e}^{ - b(t - u)} du$

$\longmapsto \rm \: = \int_0^t \: {e}^{ - au - bt + bu} du$

$\longmapsto \rm \: = \int_0^t \: {e}^{ u(b - a) - bt} du$

$\longmapsto \rm \: = \bigg( \dfrac{ {e}^{(b - a)u - bt} }{b - a} \bigg)_0^t$

$\longmapsto \rm \: = \dfrac{1}{b - a} \bigg( {e}^{(b - a)t - bt} - {e}^{ - bt} \bigg)$

$\longmapsto \rm \: = \dfrac{1}{b - a} \bigg( {e}^{- at} - {e}^{ - bt} \bigg)$