Question

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

22.$\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{1}\mathbf{+}\mathit{t}{\mathit{e}}^{\mathbf{t}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The solution for the given initial-value using the Laplace transform is $y\left(t\right)=e^t-1+\frac{1}{2!}{t}^2{e}^t$.

Step-by-step solution

Definition of Laplace transform

A specialtype of integral transform called the Laplace transform, posses the linearity property in solving linear initial-value problems.

Let f be a function defined for $t\ge 0$.

$L\left\{f\left(t\right)=\right\}=\underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt$

Then the integral is said to be the Laplace transform of f, provided that the integral converges.

We have the form in,

$y\text{'}-y=1+t{e}^t......y\left(0\right)=0$

Step 2: THEOREM 7.2.2 Transform of a Derivative

If $f,f\text{'},.....,f^{\left(n-1\right)}$are continuous onand are of exponential order and if$f^{\left(n\right)}\left(t\right)$is piecewise continuous on$[0,\infty )$, then

$L\left\{f^{\left(n\right)}\left(t\right)\right\}=s^{n}F\left(s\right)-s^{n-1}f\left(0\right)-s^{n-2}f\text{'}\left(0\right)-...-f^{(n-1)}\left(0\right)$

Where $F\left(s\right)=L\left\{f\left(t\right)\right\}$.

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

We know that $\frac{1}{{\left(s-1\right)}^3}=\underset{s\to s-1}{\lim }\frac{1}{s^3}$and $\frac{1}{s^3}=\frac{1}{2!}\frac{2!}{s^3}=\frac{1}{2!}L\left\{t^2\right\}$

The limit $\underset{s\to s-a}{\lim }$gets replaced with $e^{at}$

But, we need to decompose the fraction, $\frac{1}{s\left(s-1\right)}$

$\begin{array}{rcl}\frac{1}{s\left(s-1\right)}& =& \frac{a}{s}+\frac{b}{s-1}=\frac{1}{s-1}-\frac{1}{s}\\ y\left(t\right)& =& e^t-1+\frac{1}{2!}{t}^2{e}^t\end{array}$

THEOREM 7.3.1 First Translation Theorem

If$L\left\{f\left(t\right)\right\}=F\left(s\right)$and is any real number, then

$L\left\{e^{at}f\left(t\right)\right\}=F(s-a)$

Therefore, the solution for the given initial-value $y\text{'}-y=1+t{e}^t,y\left(0\right)=0$using the Laplace transform is $y\left(t\right)=e^t-1+\frac{1}{2!}{t}^2{e}^t$.