91Ó°ÊÓ

The equation is $y"-4y+4y=6e^{2t}$and$y_0-0,y_0-0$.

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.

$y"-4y+4y=6e^{2t}$

Take the Laplace of above equation.

$$\begin{gathered} L\left\{y"-4y\text{'}+4y\right\}=L\left\{6e^{2n}\right\}L\left\{y"\right\}-4L\left\{y\text{'}\right\}+4L\left\{y\right\} \\ =6L\left\{e^{2n}\right\}{p}^{2}L\left\{y\right\}-p{y}_0-y_0-4\left[pL\left\{y\right\}-y_0\right]+4L\left\{y\right\} \\ =\frac{6}{p-2}\left(p^2-4p+4\right)L\left\{y\right\}+\left(4-p\right)y_0-y_0^{\text{'}} \\ =\frac{6}{p-2} \end{gathered}$$

Further solve,

$$\begin{gathered} \left(p^2-4p+4\right)L\left\{y\right\}-\left(4-p\right)\left(0\right)-0=\frac{6}{p-2}\left(p^2-4p+4\right)L\left\{y\right\} \\ =\frac{6}{p-2}L\left\{y\right\} \\ =\frac{1}{p-2}\left(p^2-4p+4\right) \\ =\frac{6}{{\left(p-2\right)}^3} \end{gathered}$$

The inverse Laplace is,

$$\begin{gathered} y=L^{-1}\left\{\frac{6}{{\left(p-2\right)}^3}\right\} \\ =6L\left\{\frac{1}{{\left(p-2\right)}^3}\right\} \\ =6\frac{t^2}{2}{e}^{2t}y \\ =3t^2{e}^{2t} \end{gathered}$$

Thus, the solution of given differential equation is$y=3t^2{e}^{2t}$.