Question

Use L32 and L11 to obtain$\mathit{L}\left(t^{2}sinat\right)$.

Short answer

Answer

The given function $L\left\{t^2\sin at\right\}=\frac{2\mathrm{\pi }\left[3{\mathrm{\mu }}^2-{\mathrm{a}}^2\right]}{{\left({\sigma }^2+a^2\right)}^3}$is proved.

Step-by-step solution

Given information

The given function is$L\left\{t^2\sin at\right\}=\frac{2\mathrm{\pi }\left[3{\mathrm{\mu }}^2-{\mathrm{a}}^2\right]}{{\left({\sigma }^2+a^2\right)}^3}$

Definition of Laplace Transformation

A transformation of a function f(x) into the function g(t) 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(s) is the piecewise-continuous and exponentially-restricted real function f(t)

Properties used for prove a given function

$$\begin{gathered} L\left(32\right):L\left\{t^{n}g\left(f\right)\right\}={\left(-1\right)}^x\frac{d^{n}G\left(p\right)}{d{p}^n} \\ L\left(32\right):L\left\{t^{n}g\left(f\right)\right\}=\frac{2ap}{{\left(a^2+a^2\right)}^2} \end{gathered}$$

Proof for a given function

Now, consider the Laplace$L\left\{t\sin at\right\}$.

Use LS2 and L11.

$$\begin{gathered} L\left\{t^{n}g\left(t\right)\right\}={\left(-1\right)}^n\frac{d^n\left(3p\right)}{d{p}^n} \\ L\left\{t^1\left(t\sin at\right)\right\}={\left(-1\right)}^1\frac{d}{dp}\left[\frac{2ap}{{\left(p^2+{\alpha }^2\right)}^2}\right] \\ L\left\{t^2\sin at\right\}=-\frac{d}{dp}\left[\frac{2ap}{\left(p^2+u^2\right)}\right] \end{gathered}$$

Differentiate with quotient rule,

$\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{fg-fg}{g^2}$

$$\begin{gathered} L\left\{t^2\sin at\right\}=-\frac{d}{dp}\left[\frac{2ap}{{\left(p^2+a^2\right)}^2}\right] \\ L\left\{t^2\sin at\right\}=-\left[\frac{2a{\left(p^2+a^2\right)}^2-2\left(p^2+a^2\right)\left(2p\right)\left(2ap\right)}{{\left({\left(p^2+a^2\right)}^2\right)}^2}\right] \\ L\left\{t^2\sin at\right\}=-\frac{-2a{\left(p^2+a^2\right)}^2+8a{p}^2\left(p^2+a^2\right)}{{\left(p^2+a^2\right)}^4} \\ L\left\{t^2\sin at\right\}=-\frac{\left(p^2+a^2\right)\left|-2a\left(p^2+a^2\right)+3a{p}^2\right|}{{\left(p^2+a^2\right)}^2} \end{gathered}$$

Solving further;

$$\begin{gathered} L\left\{t^2\sin at\right\}=\frac{\left(p^2+a^2\left|-2a{p}^2-2u^3+3a{p}^2\right|\right)}{{\left(p^2+a^2\right)}^4} \\ L\left\{t^2\sin at\right\}=\frac{\left|-2a^3+6a{v}^2\right|}{{\left(p^2+a^2\right)}^3} \\ L\left\{t^2\sin at\right\}=\frac{2a\left|3p^2-a^2\right|}{{\left(p^2+a^2\right)}^3} \end{gathered}$$

Hence, $L\left\{t^2\sin at\right\}=\frac{2a\left|3p^2-a^2\right|}{{\left(p^2+a^2\right)}^3}$.