By using Laplace transforms, solve the following differential equations subject to the given initial conditions.

$\mathit{y}\mathbf{"}\mathbf{+}\mathbf{16}\mathit{y}\mathbf{=}\mathbf{8}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{4}\mathbf{}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}{\mathit{y}}_{\mathbf{0}}\mathbf{=}{\mathit{y}}_{\mathbf{0}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0}$

The given equation is $y"+16y=8\cos 4t$and$y_0-0,y_0-0$.

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)

Consider the equation.

$y"+16y=8\cos 4t$

Take the Laplace of above equation.

$$\begin{gathered} L\left\{y"+16y\right\}=L\left\{8\cos 4t\right\}L\left\{y\right\}-py0-y_0^{\text{'}}+16 \\ L\left\{y\right\}=8L\left\{\cos 4t\right\}{p}^{2}L\left\{y\right\}-py0-y_0+16 \\ L\left\{y\right\}=\frac{8p}{p^2+16}\left(p^2+16\right) \\ L\left\{y\right\}-p{y}_0-y_0^{\text{'}}=\frac{8p}{p^2+16} \end{gathered}$$

Further solve,

$$\begin{gathered} \left(p^2+16\right)L\left\{y\right\}-p\left(0\right)=\frac{8p}{p^2+16}\left(p^2+16\right) \\ L\left\{y\right\}=\frac{8p}{p^2+16} \\ L\left\{y\right\}=\frac{8p}{{\left(p^2+16\right)}^2} \end{gathered}$$

The inverse Laplace is,

$$\begin{gathered} y=L^{-1}\left\{\frac{8p}{{\left(p^2+16\right)}^2}\right\} \\ =L^{-1}\left\{\frac{2-4p}{{\left(p^2+16\right)}^2}\right\}y \\ =t\sin 4t \end{gathered}$$

Thus, the solution of given differential equation is$y=t\sin 4t$.