Question

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

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

Short answer

The solution for the given initial-value using the Laplace transform is

$y\left(t\right)=t{e}^{-4t}+2e^{-4t}$

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 $\mathit{f}$be a function defined for role="math" localid="1664228932571" $\mathit{t}\mathbf{\ge }\mathbf{0}$.

Then the integral $L\left\{f\left(t\right)=\right\}=\underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt$is said to be the Laplace transform of f, provided that the integral converges.

We have the form in,

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

Step 2: THEOREM 7.2.2 Transform of a Derivative

If$f,f\text{'},.....,f^{\left(n-1\right)}$are continuous on$[0,\infty )$and are of exponential order and ifis 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 $F\left(s\right)=L\left\{f\left(t\right)\right\}\frac{1}{n}$.

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

Now, $y\left(t\right)=L^{-1}\left\{\frac{1}{{\left(s+4\right)}^2}\right\}+L^{-1}\left\{\frac{2}{s-(-4)}\right\}$

We used to decompose the fraction, but now we have a better way to do it.

$\frac{1}{{\left(s+4\right)}^2}=\underset{s\to s-(-4)}{\lim }\frac{1}{s^2}=t{e}^{-4t}$

$y\left(t\right)=t{e}^{-4t}+2e^{-4t}$

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 $y\text{'}+4y=e^{-4t},y\left(0\right)=2$using the Laplace transform is $y\left(t\right)=t{e}^{-4t}+2e^{-4t}$.