Q12E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the convolution theorem to find the inverse Laplace transform of the given function.

$\frac{\mathbf{s}\mathbf{+}\mathbf{1}}{{\mathbf{(}{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}$

Step-by-Step Solution

Answer

The inverse Laplace transform for the given function by using the convolution theorem is.

 $y\left(t\right)=\frac{1}{2}[t\sin t-t\cos t+\sin t]$

Step 1: Define convolution theorem

Let$f\left(t\right)$and be piecewise continuous on$[0,\infty )$ and of exponential orderand

set,$F\left(s\right)=\mathcal{L}\left\{f\right\}\left(s\right)\text{ and }G\left(s\right)=\mathcal{L}\left\{g\right\}\left(s\right)$ then,

$\mathcal{L}\{f\ast g\}\left(s\right)=F\left(s\right)G\left(s\right),$or

${\mathcal{L}}^{-1}\left\{F\right(s\left)G\right(s\left)\right\}\left(t\right)=(f\ast g)\left(t\right)$

 

Step 2: Apply inverse Laplace transform and use the convolution theorem

Consider the given function,

$\frac{s+1}{{(s^2+1)}^2}$

Let,$y\left(s\right)=\frac{s+1}{{(s^2+1)}^2}$

Take inverse Laplace transform on both sides, 

${\mathcal{L}}^{-1}\left[y\right(s\left)\right]={\mathcal{L}}^{-1}\left[\frac{s+1}{{(s^2+1)}^2}\right]$$\to \left(1\right)$

Hence, the convolution formula is,${\mathcal{L}}^1\left[f\left(s\right)\cdot g\left(s\right)\right]=f\star g={\int }_0^{t}f(t-v)g\left(v\right)dv$ , where and 

$f\left(s\right)=\frac{1}{s^2+1}$And,$f\left(t\right)=\sin t$

$\begin{array}{c}g\left(s\right)=\frac{s+1}{s^2+1}\\ =\frac{s}{s^2+1}+\frac{1}{s^2+1}\end{array}$ and $g\left(t\right)=\cos t+\sin t$

Thus, equation(1) becomes,

$\begin{array}{c}y\left(t\right)={\int }_0^t\sin (t-v)\cdot [\cos v+\sin v]dv\\ ={\int }_0^t\sin (t-v)\cos vdv+{\int }_0^t\sin (t-v)\sin vdv\\ ={\int }_0^t\frac{1}{2}[\sin t+\sin (t-2v\left)\right]dv+{\int }_0^t\frac{-1}{2}[\cos t-\cos (t-2v\left)\right]dv\end{array}$

Step 3: Simplify the function using the required formula

Use the formula,$\begin{array}{c}\sin (A+B)+\sin (A-B)=2\sin A\cdot \cos B\\ \cos (A+B)-\cos (A-B)=-2\sin A\cdot \sin B\end{array}$

$\begin{array}{c}y\left(t\right)={\int }_0^t\frac{1}{2}\sin tdv+{\int }_0^t\frac{1}{2}\sin (t-2v)dv-{\int }_0^t\frac{1}{2}\cos tdv+{\int }_0^t\frac{1}{2}\cos (t-2v)dv\\ =\frac{1}{2}\sin t{\left[v\right]}_0^t+\frac{1}{4}\cos {[t-2v]}_0^t-\frac{1}{2}\cos {\left[v\right]}_0^t-\frac{1}{4}\sin (t-2v){|}_0^t\\ =\frac{1}{2}t\sin t-0-\frac{1}{2}t\cos t+\frac{1}{2}\sin t\\ =\frac{1}{2}[t\sin t-t\cos t+\sin t]\end{array}$

Therefore, the inverse Laplace transform is.$y\left(t\right)=\frac{1}{2}[t\sin t-t\cos t+\sin t]$