Question

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

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

Short answer

The required solution for the given initial-value problem by using the Laplace transform is.$y\left(t\right)=\left\{\begin{array}{l}\sin t,2m\pi ⩽t<(2m+1)\pi \\ 0(2m+1)\pi ⩽t<(2m+2)\pi \end{array}\left(m=0,1,...\right)\right.$

The graph can be plotted on the interval $[0,8\pi ]$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, 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 be obtained as,$\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\hat{\mathbf{y}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{\alpha u}\left(\mathbf{t}\mathbf{-}{\mathbf{t}}_{\mathbf{0}}\right)\mathbf{w}\left(\mathbf{t}\mathbf{-}{\mathbf{t}}_{\mathbf{0}}\right)$

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

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

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

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

Substitute the values ofin the equation$w={\mathcal{L}}^{-1}\left(\frac{1}{a{s}^2+bs+c}\right)$

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

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

$y\left(t\right)=\hat{y}\left(t\right)+\sum _{k=0}^{\infty }u(t-k\pi )w(t-k\pi )=\sin t\sum _{k=0}^{\infty }(-1{)}^{k}u(t-k\pi )$

Thus,$*y\left(t\right)=\left\{\begin{array}{l}\sin t,2m\pi ⩽t<(2m+1)\pi \\ 0(2m+1)\pi ⩽t<(2m+2)\pi \end{array}\left(m=0,1,...\right)\right.$

Therefore, the solution for the initial-value problem is

$y\left(t\right)=\left\{\begin{array}{l}\sin t,2m\pi ⩽t<(2m+1)\pi \\ 0(2m+1)\pi ⩽t<(2m+2)\pi \end{array}\left(m=0,1,...\right)\right.$.

Hence, the graph can be plotted as,