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

$\mathit{y}\mathbf{"}\mathbf{+}\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{t}\mathbf{,}\mathbf{}\mathbf{3}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}{\mathbf{\text{'}}}_{\mathbf{0}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}$,

The given equation is$y"+y=sint$and$y_0=0,y{\text{'}}_0=-\frac{1}{2}$

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.

$y"+y=\sin t$

Take the Laplace of above equation.

$$\begin{gathered} L\left\{y"+y\right\}=L\left\{\sin t\right\} \\ L\left\{y\mathit{"}\right\}+L\left\{y\right\}=L\left\{\sin t\right\}{p}^2 \\ L\left\{y\right\}-p{y}_0-y{\text{'}}_0+L\left\{y\right\}=\frac{1}{p^2+1}\left(p^2+1\right)L\left\{y\right\}-p{y}_0-y{\text{'}}_0 \end{gathered}$$

Further solve,

$$\begin{gathered} \left(p^2+1\right)L\left\{y\right\}-p\left(0\right)-\left(-\frac{1}{2}\right)=\frac{1}{p^2+1}\left(p^2+1\right) \\ L\left\{y\right\}=\frac{1}{p^2+1}\frac{1}{2} \\ L\left\{y\right\}=\frac{1}{{\left(p^2+1\right)}^2}-\frac{1}{2\left(p^2+1\right)} \end{gathered}$$

The inverse Laplace is,

$$\begin{gathered} y=L^{-1}\left\{\frac{1}{{\left(p^2+1\right)}^2}\right\}-L^{-1}\left\{\frac{1}{2{\left(p^2+1\right)}^2}\right\} \\ =-\frac{1}{2}\sin t+\frac{1}{2}\left[\sin t-t\cos t\right]y \\ =-\frac{1}{2}t\cos t \end{gathered}$$

Thus, the solution of given differential equation is $y=-\frac{1}{2}t\cos t$.