Question

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

$\mathbf{}\mathbf{3}\mathbf{.}\mathbf{}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{\delta }\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{\pi }\mathbf{)}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

The solution for the initial value problem is$y\left(t\right)=\sin (t-2\pi )U(t-2\pi )+\sin \left(t\right)$

Step-by-step solution

Define Laplace transform:

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

Find y(t)

Use Theorem 7.5.1, Transform of the Dirac Delta function

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

$y\text{'}\text{'}+y=∆(t-2\pi )\dots \dots y\left(0\right)=0\dots y\text{'}\left(0\right)=1$

$t_0=2\pi \to$

$L\left\{y\text{'}\text{'}+y\right\}=L\{∆(t-2\pi \left)\right\}\to$

$s^{2}Y-s\left(0\right)-1+Y=e^{-2\pi s}\to Y=\frac{e^{-2\pi s}}{s^2+1}+$$\frac{1}{s^2+1}$

$y\left(t\right)=L^{-1}\left\{\frac{e^{-2\pi s}}{s^2+1}+\frac{1}{s^2+1}\right\}$

Thus, the solution is

$y\left(t\right)=\sin (t-2\pi )U(t-2\pi )+\sin \left(t\right)$