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

$\mathit{y}\mathbf{"}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathit{e}\mathbf{-}\mathbf{2}\mathit{t}\mathbf{,}\mathit{y}\mathit{v}\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{\xi }}_{\mathbf{0}}\mathbf{=}\mathbf{4}$

The given equation is $y"+4y+4y+4y=e^{-2t}$and$y_0=0,=4$.

A transformation of a function f(x) into the function g(t) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function F(s) is the piecewise-continuous and exponentially-restricted real function f(t)

Consider the equation.

$y"+4y\text{'}+4y=e^{-2x}$

Take the Laplace of above equation.

role="math" localid="1654165768435" $$\begin{gathered} L\left\{y"+4y\text{'}+4y\right\}=L\left\{e^{-2t}\right\}L\left\{e^{-2t}\right\}L\left\{y"\right\}+4L\left\{y\text{'}\right\}+4L\left\{y\right\} \\ =L\left\{e^{-2t}\right\}{p}^{2}L\left\{y\right\}-p{y}_0-y_0+4\left[pL\left\{y\right\}-y_0\right]+4L\left\{y\right\} \\ =\frac{1}{p+2}\left(p^2+4p+4\right)L\left\{y\right\}-\left(p+4\right){\mathrm{y}}_0-{\mathrm{y}}_0^{\text{'}} \\ =\frac{1}{\mathrm{p}+2} \end{gathered}$$

Further solve,

role="math" localid="1654166068046" $$\begin{gathered} \left(p^2+4p+4\right)L\left\{y\right\}-\left(p+4\right)\left(0\right)-4=\left(p^2+4p+4\right) \\ L\left\{y\right\}=\frac{1}{p+2}+4 \\ L\left\{y\right\}=\frac{1}{p+2\left(p^2+4p+4\right)}+\frac{4}{\left(p^2+4p+4\right)} \\ =\frac{1}{{\left(p+2\right)}^3}+\frac{4}{{\left(p+2\right)}^2} \end{gathered}$$

The inverse Laplace is,

$$\begin{gathered} y=L^{-1}\left\{\frac{1}{{\left(p+2\right)}^3}\right\}+L^{-1}\left\{\frac{4}{{\left(p+2\right)}^2}\right\} \\ =L^{-1}\left\{\frac{1}{{\left(p+2\right)}^3}\right\}+4L^{-1}\left\{\frac{1}{{\left(p+2\right)}^2}\right\} \\ =\frac{t^2}{2I}{e}^{-2t}+4\left(\frac{t}{1I}{e}^{-2t}\right)y \\ =e^{2t}\left(\frac{t^2}{2}+4t\right) \end{gathered}$$

Thus, the solution of given differential equation is$y=e^{2t}\left(\frac{t^2}{2}+4t\right)$.