By using Laplace transforms, solve the following differential equations subject to the given initial conditions.

$\ddot{\mathbf{y}}\mathbf{+}\mathbf{4}\dot{\mathbf{y}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{2}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{t}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{,}{\dot{\mathbf{y}}}_{\mathbf{0}}\mathbf{=}\mathbf{3}$

The differential equation is $\ddot{y}+4\dot{y}+5y=2e^{-2t}cost,y_0=0,{\dot{y}}_0=3$.

Application of Laplace transform:

It's used to simplify complicated differential equations by using polynomials. It's used to convert derivatives into numerous domain variables, then utilise the Inverse Laplace transform to convert the polynomials back to the differential equation.

$L\left(2e^{-at}\cos t\right)$The initial conditions $y_0=0,{\dot{y}}_0=3$.

Consider L35 result of Laplace transforms of derivative of y

$$\begin{gathered} L\left(y\right)=Y \\ L\left(\dot{y}\right)=pY-y_0 \\ L\left(\ddot{y}\right)=p^{2}Y-p{y}_0-{\dot{y}}_0 \\ L\left(e^{-at}\cos bt\right)=\frac{p+a}{{\left(p+a\right)}^2+b^2},\mathrm{Re}\left(p+a\right)>\left|\mathrm{lm}b\right| \end{gathered}$$

Use Laplace on both sides of given differential equation

$L\left(\ddot{y}+4\dot{y}+5y\right)=L\left(2e^{-2t}\cos t\right)$

Compare right hand side term $L\left(2e^{-2t}\cos t\right)$with the general term .

a = 2,b = 1

Use linearity property of Laplace transformation

$$\begin{gathered} L\left(\ddot{y}\right)+4L\left(\dot{y}\right)+5L\left(y\right)=2L\left(e^{-2t}\cos t\right) \\ \left(p^{2}y-p{y}_0-\dot{y_0}\right)+4\left[pY-y_0\right]+5y=2\frac{p+2}{{\left(p+2\right)}^2+1^2} \\ \left(p^2+4p+5\right)Y=\frac{2\left(p+2\right)}{{\left(p+2\right)}^2+1}3 \end{gathered}$$

Consider the expression

$$\begin{gathered} p^2+4p+5=p^2+4p+4+1 \\ ={\left(p+2\right)}^2+1^2 \end{gathered}$$

Use this in above equation

$$\begin{gathered} \left(p^2+4p+5\right)Y=\frac{2\left(p+2\right)}{{\left(p+2\right)}^2+1}+3 \\ y=\frac{2\left(p+2\right)}{\left[{\left(p+2\right)}^2+1^2\right]\left[{\left(p+2\right)}^2+1\right]}+\frac{3}{{\left(p+2\right)}^2+1^2} \\ y=\frac{2\left(p+2\right)}{\left[{\left(p+2\right)}^2+1^2\right]}+\frac{3}{{\left(p+2\right)}^2+1^2} \end{gathered}$$

The first term $\frac{2(p+2)}{{\left[{\left(p+2\right)}^2+1^2\right]}^2}$contains variable (p + 2) instead of p.

Use frequency shift theorem to replace p with (p - 2). From this replacement, one can use Laplace inverse transformation.

The second term $\frac{3}{(p+2{]}^2+1^2}$is comparable with $\sin at=L^{-1}\left[\frac{a}{p^2+a^2}\right]$.

The frequency shift theorem is

$$\begin{gathered} y\left(t\right)=L^{-1}\left\{Y\left(p\right)\right\} \\ \left\{e^{-p_{0}t}y\left(t\right)\right\}=L^{-1}\left\{Y\left(p+p_0\right)\right\} \end{gathered}$$

Substitute -2 for p and (p - 2) for p in above equation

$\left\{e^{-\left(-2\right)t}y\left(t\right)\right\}=L^{-1}\left\{Y\left(p+\left(2\right)\right)\right\}$

Solve further

$e^{2t}y\left(t\right)=L^{-1}\left\{Y\left(p-2\right)\right\}$ ……. (1)

$Y\left(p\right)=\frac{2\left(p+2\right)}{{\left[{\left(p+2\right)}^2+1^2\right]}^2}+\frac{3}{{\left(p+2\right)}^2+1^2}$

Substitute p for p-2 in above expression.

role="math" localid="1659252776741" $$\begin{gathered} Y\left(p-2\right)=\frac{2\left(p-2+2\right)}{\left[{\left(p-2+2\right)}^2+1^2\right]}+\frac{3}{{\left(p-2+2\right)}^2+1^2} \\ =\frac{2\left(p\right)}{{\left[{\left(p\right)}^2+1^2\right]}^2}+\frac{3}{{\left(p\right)}^2+1^2} \end{gathered}$$

Substitute the values in equation (1)

$$\begin{gathered} e^{2t}\left(t\right)=L^{-1}\left\{Y\left(p-2\right)\right\} \\ {e}^{2t}\left(t\right)=L^{-1}\left[\frac{2\left(p\right)}{{\left[{\left(p\right)}^2+1^2\right]}^2}+\frac{3}{p^2+1^2}\right] \\ {e}^{2t}\left(t\right)=L^{-1}\left(\frac{2·1·p}{\left(p^2+1^2\right)}\right)+3L^{-1}\left(\frac{1}{p^2+1^2}\right) \end{gathered}$$

Further solve

$\begin{array}{l}{e}^{2t}y\left(t\right)=t\sin +3\sin t\\ e^{2t}y\left(t\right)=\left(t+3\right)\sin t\end{array}$

Divide both side by

$y\left(t\right)=e^{-2t}(t+3)\sin t$

Thus, the given differential equation's solution is $y\left(t\right)=e^{-2t}(t+3)\sin t$.