Question

Solve Laplace transforms and the convolution integral or by Green functions.

$\mathbf{y}\mathbf{\text{'}\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{t}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{t}$

Short answer

The solution of given equation is$y\left(t\right)=\frac{t\sin t}{4}-\frac{t^2\cos t}{4}$

Step-by-step solution

Given Information

The expression is$y\text{'}\text{'}+y=t\sin t$

Green Function

The impulse response of an inhomogeneous linear differential operator defined on a domain with specified initial conditions or boundary conditions is known as a Green's function.

Laplace Transform

Consider the equation.

$y\text{'}\text{'}+y=t\sin t$

Take Laplace transform of both side of above equation.

$L\left(y\text{'}\text{'}+y\right)=L\left(t\sin t\right)$

$$\begin{gathered} p^{2}L\left(y\right)+L\left(y\right)=\frac{2p}{{\left(p^2+1\right)}^2} \\ L\left(y\right)=\frac{2p}{\left(p^2+1\right){\left(p^2+1\right)}^2} \\ y\left(t\right)=L^{-1}\left\{\frac{2P}{{\left(p^2+1\right)}^2}\frac{1}{\left(p^2+1\right)}\right\} \\ y\left(t\right)={\int }_0^t\tau \sin \tau \sin \left(t-\tau \right)d\tau \\ ={\int }_0^t\tau \sin \tau \left(\sin t\cos \tau -\cos t\sin \tau \right)d\tau \\ =\frac{\sin t}{t}{\left[\tau \left(\frac{-\cos 2\tau }{2}\right)+\frac{\sin 2\tau }{2}\right]}_0^t-\frac{\cos t}{2}{\left[\frac{{\tau }^2}{2}\right]}_0^t+\frac{\cos t}{2}{\left[\frac{\cos 2\tau }{4}+\tau \frac{\sin 2\tau }{2}\right]}_0^t \\ y\left(t\right)=\frac{t\sin t}{4}-\frac{t^2\cos t}{4} \end{gathered}$$

Thus, the solution of given equation is $y\left(t\right)=\frac{t\sin t}{4}-\frac{t^2\cos t}{4}$.