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

$\mathit{y}\mathbf{"}\mathbf{+}\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{t}\mathbf{}\mathbf{}\mathbf{}\mathbf{,}$

The given equation is $y^{"}+y=\sin t$and$y_0=1y_0=0$.

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.

$y^{"}+y=\sin t$

Take the Laplace of above equation.

$$\begin{gathered} L\left\{y^{"}+y\right\}=L\left\{\sin t\right\}L\left\{y^{"}\right\}+L\left\{y\right\} \\ =L\left\{\sin t\right\}{p}^{2}L\left\{y\right\}-p{y}_0+L\left\{y\right\} \\ =\frac{1}{p^2+1}\left(p^2+1\right)L\left\{y\right\}-p{y}_0-y_0 \\ =\frac{1}{p^2+1} \end{gathered}$$

Further solve,

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

The inverse Laplace is,

$$\begin{gathered} y=L^{-1}\left\{\frac{1}{{\left(p^2+1\right)}^2}\right\}+L^{-1}\left\{\frac{p}{\left(p^2+1\right)}\right\} \\ =\frac{1}{2{\left(1\right)}^3}\left[\sin \left(1\right)t-t\cos \left(1\right)t\right]+\cos \left(1\right)ty \\ =\frac{1}{2}\left(\sin t-t\cos t\right)+\cos t \end{gathered}$$

Thus, the solution of given differential equation is$y-\frac{1}{2}\left(\sin t-t\cos t\right)+\cos t$.