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

$\mathit{y}\mathbf{"}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{4}$, $\mathbf{\sum }_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{\psi }}_{\mathbf{0}}\mathbf{=}\mathbf{-}\mathbf{2}$

The given equation is $y"-4y\text{'}+4y=4$and$y_0=0,y_n=-2$.

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.

$y"-4y\text{'}+4y=4$

Take the Laplace of above equation.

$$\begin{gathered} L\left\{y"-4y\text{'}+4y\right\}=L\left\{4\right\}L\left\{y"\right\}-4L\left\{y\text{'}\right\}+4L\left\{y\right\} \\ =L\left\{4\right\}{p}^{2}L\left\{y\right\}-p{y}_0-y_0^{\text{'}}-4\left[pL\left\{y\right\}-y_0\right]+4L\left\{y\right\} \\ =\frac{4}{p}\left(p^2-4p+4\right)L\left\{y\right\}+\left(4-p\right)y_0-y_0^{\text{'}} \\ =\frac{4}{p} \end{gathered}$$

Further solve,

$$\begin{gathered} \left(p^2-4p+4\right)L\left\{y\right\}+\left(4-p\right)\left(4-p\right)\left(0\right)-\left(-2\right)=\frac{4}{p}\left(p^2-4p+4\right) \\ L\left\{y\right\}=\frac{4}{p}-2 \\ L\left\{y\right\}=\frac{4}{p\left(p^2-4p+4\right)}-\frac{2}{\left(p^2-4p+4\right)} \\ =\frac{4}{p{\left(p-2\right)}^2}-\frac{2}{{\left(p-2\right)}^2} \end{gathered}$$

The inverse Laplace is,

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

Thus, the solution of given differential equation is$y=1-e^{2t}$.