Question

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

23. $\mathit{y}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

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

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

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 $t\ge 0$.

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,

$y+2y\text{'}+y=0,y\left(0\right)=1,y\text{'}\left(0\right)=1$

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

Transforming derivatives:

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

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

$L\left\{y+2y\text{'}+y\right\}=L\left\{0\right\}\to s^{2}Y-s\left(1\right)-1+2\left(sY-1\right)+Y=0$

$\Rightarrow Y=\frac{s+3}{\begin{array}{l}{s}^2+2s+1\\ =\frac{s+3}{{\left(s+1\right)}^2}\end{array}}$

Now we need to decompose, $\frac{s+3}{{\left(s+1\right)}^2}$

$\begin{array}{l}\frac{s+3}{{\left(s+1\right)}^2}\Rightarrow \frac{a}{s+1}+\frac{b}{{\left(s+1\right)}^2}=\frac{1}{s+1}+\frac{2}{{\left(s+1\right)}^2}\\ \end{array}$

To find the function, inverse the transform,

$\begin{array}{rcl}{L}^{-1}\left\{\frac{1}{s-(-1)}\right\}+2L^{-1}\left\{\frac{1}{{\left(s-(-1)\right)}^2}\right\}& =& e^{-t}+2L^{-1}\left\{\underset{s\to s-(-1)}{\lim }\frac{1}{s^2}\right\}\\ y\left(t\right)& =& e^{-t}+2t{e}^{-t}\end{array}$

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 $y+2y\text{'}+y=0,y\left(0\right)=1,y\text{'}\left(0\right)=1$ using the Laplace transform is$y\left(t\right)=e^{-t}+2t{e}^{-t}$.