Question

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

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{13}\mathit{y}\mathbf{=}\mathit{\delta }\mathbf{(}\mathit{t}\mathbf{-}\mathit{\pi }\mathbf{)}\mathbf{+}\mathit{\delta }\mathbf{(}\mathit{t}\mathbf{-}\mathbf{3}\mathit{\pi }\mathbf{)}\mathbf{}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The solution for the initial-value problem by using the Laplace transform is$y\left(t\right)=\frac{1}{3}{e}^{-2(t-\pi )}\sin \left[3\right(t-\pi \left)\right]U(t-\pi )+\frac{1}{3}{e}^{-2(t-3\pi )}\sin \left[3\right(t-3\pi \left)\right]U(t-3\pi )+e^{-2t}\cos \left(3t\right)+\frac{2}{3}{e}^{-2t}\sin \left(3t\right)$

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.

Determine the required solution by using the Dirac Delta Function theorem

Applying Laplace operator in both sides of equation, $y\text{'}\text{'}+4y\text{'}+13y=\delta (t-\pi )+\delta (t-3\pi ),,y\left(0\right)=1,y\text{'}\left(0\right)=0$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:

$\mathcal{L}\left\{y\text{'}\text{'}+4y\text{'}+13y\right\}=\mathcal{L}\left\{\delta \left(t-\pi \right)+\delta \left(t-3\pi \right)\right\}$

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

$$\begin{gathered} s^{2}Y-s\left(1\right)-0+4\left(sY-1\right)+13Y=e^{-\pi s}+e^{-3\pi s} \\ Y\left(s\right)=\frac{e^{-\pi s}}{s^2+4s+13}+\frac{e^{-3\pi s}}{s^2+4s+13}+\frac{s}{s^2+4s+13}+\frac{4}{s^2+4s+13} \\ Y=\frac{e^{-\pi s}}{{(s+2)}^2+9}+\frac{e^{-3\pi s}}{{(s+2)}^2+9}+\frac{s+2}{{(s+2)}^2+9}+\frac{2}{{(s+2)}^2+9} \end{gathered}$$

Take inverse Laplace Transform

For evaluating y(t), utilize inverse form of second translation theorem as:

$y\left(t\right)=L^{-1}\left\{Y=\frac{e^{-\pi s}}{{\left(s+2\right)}^2+9}+\frac{e^{-3\pi s}}{{\left(s+2\right)}^2+9}+\frac{s+2}{{\left(s+2\right)}^2+9}+\frac{2}{{\left(s+2\right)}^2+9}\right\}$

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

$y\left(t\right)=\frac{1}{3}{e}^{-2(t-\pi )}\sin \left[3\right(t-\pi \left)\right]U(t-\pi )+\frac{1}{3}{e}^{-2(t-3\pi )}\sin \left[3\right(t-3\pi \left)\right]U(t-3\pi )+e^{-2t}\cos \left(3t\right)+\frac{2}{3}{e}^{-2t}\sin \left(3t\right)$Simplify the equation using $\sin (t-\pi )=\sin (t-3\pi )=-\sin \left(t\right)$,

.$y\left(t\right)=-\frac{1}{3}{e}^{-2(t-\pi )}\sin \left(t\right)U(t-\pi )-\frac{1}{3}{e}^{-2(t-3\pi )}\sin \left(t\right)U(t-3\pi )+e^{-2t}\cos \left(3t\right)+\frac{2}{3}{e}^{-2t}\sin \left(3t\right)$