Question

In Problems 21–30 use the Laplace transform to solve the given initial-value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}{\mathit{t}}^{\mathbf{3}}{\mathit{e}}^{\mathbf{2}\mathbf{t}}\mathbf{,}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{ }{\mathit{y}}^{\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{0}$

Short answer

Transforming derivatives

$$\begin{gathered} L\left\{y^{\text{'}\text{'}}\right\}=s^{2}Y-sy\left(0\right)-y^{\text{'}}\left(0\right), \\ L\left\{y^{\text{'}}\right\}=sY-y\left(0\right) \\ Y\left(t\right)=\frac{6}{5!}{t}^5{e}^{2t}=\frac{1}{20}{t}^5{e}^{2t} \end{gathered}$$

Step-by-step solution

Definition of Laplace transform

Laplace transform is the integral transform of the given derivative function with real variable t to convert into a complex function with variables.

$y^{\text{'}\text{'}}-4y^{\text{'}}+4y=t^3{e}^{2t}...y\left(0\right)=0, y^{\text{'}}\left(0\right)=0$

Transforming derivatives

Transforming derivatives (Theorem 7.2.2):

$L\left\{y^{\text{'}\text{'}}\right\}=s^{2}Y-sy\left(0\right)-y^{\text{'}}\left(0\right),$

And,$L\left\{y^{\text{'}}\right\}=sY-y\left(0\right)$

$L\left\{y^{\text{'}\text{'}}-4y^{\text{'}}+4y\right\}=L\left\{t^3{e}^{2t}\right\}\to$

Now, the function of theorem is,

$s^{2}Y-s\left(0\right)-1\left(0\right)-4\left(sY-0\right)+4Y=\frac{3!}{{\left(s-2\right)}^4}\to Y=\frac{6}{{\left(s-2\right)}^2{\left(s-2\right)}^4}=\frac{6}{{\left(s-2\right)}^6}$

Use the theorem

Now we don't need to decompose $\frac{6}{{\left(s-2\right)}^6}$,

Then, we will use Theorem,

$\frac{6}{{\left(s-2\right)}^6}=6 \underset{s\to s-2}{\lim }\frac{1}{5!}\frac{5!}{s^6}$

Now simplify this,

$L^{-1}\left\{6\underset{s\to s-2}{\lim } \frac{1}{5!}\frac{1}{s^6}\right\}$

Here Result is,

$Y\left(t\right)=\frac{6}{5!}{t}^5{e}^{2t}=\frac{1}{20}{t}^5{e}^{2t}$