Question

In Problems 1–12 use the Laplace transform to solve the given initial value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathit{\delta }\mathbf{(}\mathit{t}\mathbf{-}\mathbf{2}\mathit{\pi }\mathbf{)}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The required solution is$y\left(t\right)=e^{-2(t-2\pi )}\sin \left(t\right)U(t-2\pi )$

Step-by-step solution

Define Laplace Transform

The use of Laplace transformation is to convertdifferential equationsinto algebraic equations. The formula for Laplace transform is$\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{+}\mathbf{\infty }}\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{·}{\mathit{e}}^{\mathbf{-}\mathbf{s}\mathbf{·}\mathbf{t}}\mathbf{·}\mathit{d}\mathit{t}$ ,Where, F(s) is the Laplace TransformS is the complex numbert is time and is a real number.

Solve using Laplace Transform 

Applying Laplace operator in both sides of equation$y\text{'}\text{'}+4y\text{'}+5y=\delta \left(t-2\pi \right)$ using the transform$\mathcal{L}\left\{f^n\left(t\right)\right\}=s^{n}F\left(s\right)-s^{n-1}f\left(0\right)-s^{n-2}f\text{'}\left(0\right)-\dots s{f}^{n-2}\left(0\right)-f^{n-1}\left(0\right)$ as follows:

$\mathcal{L}\left\{y\text{'}\text{'}+4y\text{'}+5y\right\}=\mathcal{L}\left\{\delta \left(t-2\pi \right)\right\}$

Apply the property of Dirac delta function,$\mathcal{L}\left\{\delta \left(t-t_0\right)\right\}=e^{-s{t}_0}$, where$t_0=2\pi$ , and the given initial conditions,to write the Laplace of the above equation as follows:

$$\begin{gathered} s^{2}Y-s\left(0\right)-0+4(sY-0)+5Y=e^{-2\pi s} \\ Y\left(s\right)=\frac{e^{-2\pi s}}{s^2+4s+5} \\ =\frac{e^{-2\pi s}}{{(s+2)}^2+1} \end{gathered}$$

Take inverse Laplace Transform

Take inverse Laplace, using the formula$L\left\{e^{at}f\left(t\right)\right\}=F(s-a)$

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

Simplifying the above equation, using$\sin (t-2\pi )=\sin \left(t\right)$we obtain,

$y\left(t\right)=e^{-2(t-2\pi )}\sin \left(t\right)U(t-2\pi )$

So, the solution is.

$y\left(t\right)=e^{-2(t-2\pi )}\sin \left(t\right)U(t-2\pi )$