Q7.3 - 23E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use Theorem 4 on page362 to show how entry 32 follows from entry 31 in the Laplace transform table on the inside back cover of the text.

Step-by-Step Solution

Answer

It is proved that, $\mathcal{L}\{t\sin bt+bt\cos bt\}=\frac{2b{s}^2}{{\left(s^2+b^2\right)}^2}$ from the Laplace transform table.

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, $\mathcal{L}\left\{f\right(t\left)\right\}=F\left(s\right)$.

Show that L { t s i n b t + b t c o s b t } = 2 b s 2 s 2 + b 2 2

Consider the expression $\mathcal{L}\{t\sin bt+bt\cos bt\}$

Let $f\left(t\right)=t\sin bt$

Then, $f\text{'}\left(t\right)=t\sin bt+bt\cos bt$

Find, $\mathcal{L}\{t\sin bt+bt\cos bt\}$ using $\mathcal{L}\left\{f\text{'}\right\}\left(s\right)=s\mathcal{L}\left\{f\right\}\left(s\right)-f\left(0\right)$ and $\mathcal{L}\left\{t\sin bt\right\}=\frac{2bs}{{\left(s^2+b^2\right)}^2}$ as:

$\begin{array}{rcl}\mathcal{L}\{t\sin bt+bt\cos bt\}& =& s\mathcal{L}\left\{\right(t\sin bt\left)\right\}-f\left(0\right)\\ & =& s\frac{2bs}{{\left(s^2+b^2\right)}^2}-0\\ & =& \frac{2b{s}^2}{{\left(s^2+b^2\right)}^2}\end{array}$

Hence, it is proved that $\mathcal{L}\{t\sin bt+bt\cos bt\}=\frac{2b{s}^2}{{\left(s^2+b^2\right)}^2}$.