Question

Use the Laplace transform to solve the given initial- value problem

$\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{16}\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathit{\delta }\mathbf{}\mathbf{(}\mathit{t}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}\mathbf{\pi }\mathbf{}\mathbf{)}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The solution for the initial value problem is$y\left(t\right)=\frac{1}{4}\sin 4(t-2\pi )u(t-2\pi )$

Step-by-step solution

Define Laplace transform:

The conversion of the function f (x) to the function$g\left(t\right)=\int \infty 0e-xtf\left(x\right)dx$ is particularly useful in reducing the solution of a standard dividing line equation with constant coefficients in the polynomial equation solution

Find y(t):

Using Laplace transform we get the differential equation subject to indicated initial conditions

$y\text{'}\text{'}+16y=\delta (t-2\pi )y\left(0\right)=0y\text{'}\left(0\right)=0$

Use Laplace transform on both side

$\mathcal{L}\left\{y\text{'}\right\}+16\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{\delta \right(t-2\pi \left)\right\}$

$\mathcal{L}\left\{f^n\left(t\right)\right\}=s^{n}F\left(s\right)-s^{n-1}f\left(0\right)-s^{n-2}f\text{'}\left(0\right)-\dots s{f}^{n-2}\left(0\right)-f^{n-1}\left(0\right)$

Use Theorem 7.5.1, Transform of the Dirae Delta function

For$t_0>0$ ,$\mathcal{L}\left\{\delta \left(t-t_0\right)\right\}=e^{-2t_0}.$in above equation

${\int }_0^{\infty }f\left(t\right)\delta \left(t-t_0\right)=f\left(t_0\right)$

$\left[s\text{'}\text{'}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]+16Y\left(s\right)=e^{-2s\pi }$

Apply initial condition,

$Y\left(s\right)=\frac{e^{-2s\pi }}{s^2+4^2}$

Useutilize inverse form of second translation theorem

${\mathcal{L}}^{-1}\left\{Y\right(s\left)\right\}=\frac{1}{4}{\mathcal{L}}^{-1}\left\{\frac{4e^{-2s\pi }}{s^2+4^2}\right\}$

Thus, the solution is

$y\left(t\right)=\frac{1}{4}\sin 4(t-2\pi )u(t-2\pi )$