Question

By differentiating the appropriate formula with respect to, verify L12.

$L\left(t\cos at\right)=\frac{p^2-a^2}{{\left(p^2+a^2\right)}^2}$

Answer

Short answer

It is verified that

Step-by-step solution

Given information

The given function is$L\left(t\cos at\right)=\frac{p^2-a^2}{{\left(p^2+a^2\right)}^2}$

Definition of Laplace Transformation

A transformation of a function $f\left(x\right)$into the function $g\left(t\right)$that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function$F\left(s\right)$is the piecewise-continuous and exponentially-restricted real function$f\left(t\right)$.

Differentiate the given function

Let us consider the formula.

$$\begin{gathered} L\left(\sin at\right)={\int }_0^w{e}^{-pt}\sin atdt \\ =\frac{a}{p^2+a^2} \end{gathered}$$

Differentiate both sides with respect to 'a ' as,

${\int }_0^{=}{e}^{-pt}\left(t\cos at\right)dt=\frac{d}{du}\left[\frac{a}{p^2+a^2}\right]$

Use quotient rule, $\frac{d}{da}\left(\frac{f}{g}\right)=\frac{f^{\text{'}}g-f{g}^{\text{'}}}{g^2}$

$$\begin{gathered} {\int }_0^{\pi }{e}^{-pt}\left(t\cos at\right)dt=\frac{d}{da}\left(\frac{a}{p^2+a^2}\right) \\ =\frac{\frac{dx}{dx}\left(p^2+a^2\right)-2a\left(a\right)}{{\left(p^2+a^2\right)}^2} \\ =\frac{p^2+a^2-2a^2}{{\left(p^2+a^2\right)}^2} \\ =\frac{p^2-a^2}{{\left(p^2+a^2\right)}^2} \end{gathered}$$

Therefore,

Hence it is verified that$L\left(t\cos at\right)=\frac{p^2-a^2}{{\left(p^2+a^2\right)}^2}$