Question

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

$\mathbf{}\mathbf{5}\mathbf{.}\mathbf{}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{\delta }\left(\mathbf{t}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\pi }\right)\mathbf{+}\mathbf{\delta }\left(\mathbf{t}\mathbf{-}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{\pi }\right),\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} ,\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The solution for the initial value problem is$y\left(t\right)=\sin \left(t-\frac{\pi }{2}\right)U\left(t-\frac{\pi }{2}\right)+\sin \left(t-\frac{3\pi }{2}\right)U\left(t-\frac{3\pi }{2}\right)$

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):

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

$y\text{'}\text{'}+y=∆\left(t-\frac{1}{2}\pi \right)+∆\left(t-\frac{3}{2}\pi \right)\dots \dots y\left(0\right)=$$0\dots y\text{'}\left(0\right)=0$

We can confess,

$t_0=\left\{\frac{1}{2}\pi ,\frac{3}{2}\pi \right\}\to$

$L\left\{y\text{'}\text{'}+y\right\}=L\left\{∆\left(t-\frac{1}{2}\pi \right)+∆\left(t-\frac{3}{2}\pi \right)\right\}\to$

$s^{2}Y-s\left(0\right)-0+Y=e^{-\pi s/2}+e^{-3\pi s/2}\to Y=$$\frac{e^{-\pi s/2}}{s^2+1}+\frac{e^{-3\pi s/2}}{s^2+1}$

$y\left(t\right)=L^{-1}\left\{\frac{e^{-\pi s/2}}{s^2+1}+\frac{e^{-3\pi s/2}}{s^2+1}\right\}$

Thus, the solution is

$y\left(t\right)=\sin \left(t-\frac{\pi }{2}\right)U\left(t-\frac{\pi }{2}\right)+\sin \left(t-\frac{3\pi }{2}\right)U(t-$

We can simplify or leave as it is,

By simplifying,

$y\left(t\right)=\cos \left(t\right)\left[U\left(t-\frac{3\pi }{2}\right)-U\left(t-\frac{\pi }{2}\right)\right]$ Hint: $\sin \left(\frac{\pi }{2}-t\right)=$ $\cos \left(t\right)$