Q7.3 - 21E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Given that $\mathcal{L}\mathbf{\left\{}\mathbf{cosbt}\mathbf{\right\}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}\mathbf{s}\mathbf{/}\left(s^2+b^2\right)$, use the translation property to compute $\mathcal{L}\left\{e^{\mathrm{at}}\mathrm{cosbt}\right\}$.

Step-by-Step Solution

Answer

The value of $\mathcal{L}\left\{{\mathrm{e}}^{\mathrm{at}}\mathrm{cosbt}\right\}$ is $\frac{\mathrm{s}-\mathrm{a}}{{(\mathrm{s}-\mathrm{a})}^2+{\mathrm{b}}^2}$.

Define Laplace transform

When specific initial conditions are supplied, especially when the initial values are zero, the Laplace transform is a handy method of solving certain types of differential equations. Laplace transform $\mathcal{L}$of a function f(t) is defined as:

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^{\frac{<}{>}}{e}^{-st}f\left(t\right)dt$

In words, we can describe this expression as the Laplace transform of f(t) equals function F of s, that is, $\mathcal{L}\left\{f\right(t\left)\right\}=F\left(s\right)$.

 

Find the value of L e a t c o s b t

Given that  $\mathcal{L}\left\{\cos bt\right\}\left(s\right)=s/\left(s^2+b^2\right)$,

Find $\mathcal{L}\left\{e^{at}\cos bt\right\}\left(s\right)$ using to translation property $\mathcal{L}\left\{e^{at}f\left(t\right)\right\}\left(s\right)=F(s-a)$ as:

$\begin{array}{rcl}\mathcal{L}\left\{e^{at}\cos bt\right\}\left(s\right)& =& F(s-a)\\ & =& \mathcal{L}\left\{\cos bt\right\}(s-a)\\ & =& \frac{s-a}{{(s-a)}^2+b^2}\\ & & \end{array}$

Hence, the value of $\mathcal{L}\left\{e^{at}\cos bt\right\}$ is $\frac{s-a}{{(s-a)}^2+b^2}$.