91Ó°ÊÓ

The equation is $\mathrm{y}"+\mathrm{y}=5\sinh 2\mathrm{t}\mathrm{and}{\mathrm{y}}_0=0,{\mathrm{y}}_0^{\text{'}}=2.$

A transformation of a function $\mathbf{f}\left(x\right)$into the function $\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$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 $\mathbf{}\mathbf{F}\mathbf{\left(}\mathbf{S}\mathbf{\right)}$ is the piecewise-continuous and exponentially-restricted real function$\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$

Consider the equation.

$y"+y=5sinh2t$

Take the Laplace of above equation.

role="math" localid="1659261068654" $$\begin{gathered} L\left\{y"+y\right\}=L\left\{5\sin h2t\right\}L\left(y"\right)+L\left(y\right) \\ =5L\left\{\sin h2t\right\}{p}^{2}L\left(y\right)-p{y}_0-y{\text{'}}_0+L\left(y\right) \\ =\frac{10}{p^2-4}\left(p^2+1\right)L\left\{y\right\}-py0-y{\text{'}}_0 \\ =\frac{10}{p^2-4} \end{gathered}$$

Further solve,

role="math" localid="1659258897229" $$\begin{gathered} \left({\mathrm{p}}^2+1\right)\mathrm{L}\left\{\mathrm{y}\right\}-p\left(0\right)-\left(2\right)=\frac{10}{{\mathrm{p}}^2-4}\left({\mathrm{p}}^2+1\right)L\left\{y\right\} \\ =\frac{10}{{\mathrm{p}}^2-4}2L\left\{y\right\} \\ =\frac{10}{{\mathrm{p}}^2-4{\left({\mathrm{p}}^2+1\right)}^2}+\frac{2}{\left({\mathrm{p}}^2+1\right)}L\left\{y\right\} \\ =\frac{2}{\left({\mathrm{p}}^2-2^2\right)} \end{gathered}$$

$$\begin{gathered} \mathrm{y}={\mathrm{L}}^{-1}\left\{\frac{1}{({\mathrm{p}}^2-2^2)}\right\} \\ \mathrm{y}=\sinh 2\mathrm{t} \end{gathered}$$

Thus, the solution of given differential equation is $y=\sin h2t$.