Question

Use the Laplace transform to solve the given initial-value problem. Graph your solution on the interval $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{8}\mathit{\pi }\mathbf{]}$.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\underset{\mathbf{k}\mathbf{=}\mathbf{1}}{\overset{\stackrel{\mathbf{x}}{\mathbf{\circ }}}{\mathbf{a}}}\mathbf{\delta }\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{k\pi }\mathbf{)}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

The required solution for the initial-value problem by using the Laplace transform is.$If2m\pi ⩽t<2(m+1)\pi ,theny=(m+1)\sin t$

The graph can be plotted as

Step-by-step solution

Define the solution of initial-value problem:

Consider $\mathit{\alpha }$is a nonzero constant and the functionis continuous on the interval$\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$, then the solution of initial-value problem for${\mathbf{t}}_{\mathbf{0}}\mathbf{>}\mathbf{0}$is,

$\mathbf{ay}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{by}\mathbf{\text{'}}\mathbf{+}\mathbf{cy}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{\alpha \delta }\left(\mathbf{t}\mathbf{-}{\mathbf{t}}_{\mathbf{0}}\right),\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{k}}_{\mathbf{0}}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{k}}_{\mathbf{1}}$

Hence, it can defined as,$\mathit{y}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\hat{\mathbf{y}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{+}\mathit{\alpha }\mathit{u}\left(t-t_0\right)\mathit{w}\left(t-t_0\right)$

$\hat{\mathbf{y}}$is a solution of the equation$\mathbf{ay}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{by}\mathbf{\text{'}}\mathbf{+}\mathbf{cy}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{k}}_{\mathbf{0}}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{k}}_{\mathbf{1}}$

Therefore, $\mathbf{w}\mathbf{=}{\mathbf{L}}^{\mathbf{-}\mathbf{1}}\left(\frac{\mathbf{1}}{{\mathbf{as}}^{\mathbf{2}}\mathbf{+}\mathbf{bs}\mathbf{+}\mathbf{c}}\right)$

Solve the equation by using Laplace transform and graph the solution:

Thus, from the given differential equation, the value of$a=1,b=0andc=1$

$\hat{y}=w={\mathcal{L}}^{-1}\left(\frac{1}{s^2+1}\right)=\sin t\left({\mathcal{L}}^{-1}\left(\frac{k}{s^2+k^2}\right)=\sin kt\right)$

Hence, from the given differential equation, the equation can be written as,

Therefore,$If2m\pi ⩽t<2(m+1)\pi ,theny=(m+1)\sin t$

Hence, the solution for the graph on the interval$[0,8\pi ]$can be plotted as,