Question

In problemsanduse the Laplace transform And the procedure outlined in exampleto solve the given boundary-value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{20}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathit{\pi }\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The Laplace transform And the procedure is boundary-value problem.

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

Take Laplace transform of both side of the given equation.

Y’ (0)=c

role="math" localid="1664303766309" $$\begin{gathered} 0=L\left\{y\text{'}\text{'}+8y\text{'}+20y\right\} \\ y\left(t\right)=0 \end{gathered}$$

Step-by-step solution

definition of Laplace transform

A transformation of a function is particularly useful in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation..

give the equation:

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

Take Laplace transform of both side of the given equation.

Y’(0)=c

$$\begin{gathered} 0=L\left\{y\text{'}\text{'}+8y\text{'}+20y\right\} \\ =L\left\{y\right\}+8L\left\{y\text{'}\right\}+20L\left\{y\right\} \\ =s^{2}L\left\{y\right\}-sy\left(0\right)-y\text{'}\left(0\right)+8sL\left\{y\right\}-8y\left(0\right)+20L\left\{y\right\} \\ =\left(s^2+8s+20\right)L\left\{y\right\}-c \end{gathered}$$

Step 2: Apply inverse Laplace transform

L{Y } =Y(S) and replace in the above equation.

Addition the values:

$$\begin{gathered} 0=\left(s^2+8s+20\right)Y\left(s\right)-c \\ \begin{array}{c}c=\left(s^2+8s+20\right)Y\left(s\right)\\ Y\left(s\right)=\frac{c}{s^2+8s+20}\end{array} \end{gathered}$$

To find Y (t) we apply inverse Laplace transform to .Y(S)

Addition:

$\begin{array}{l}y\left(t\right)=L^{-1}\left\{Y\left(s\right)\right\}\\ =L^{-1}\frac{c}{s^2+8s+20}\\ =L^{-1}\frac{c}{{(s+4)}^2+4}\\ =\frac{c}{2}{L}^{-1}{\overline{)\frac{2}{s^2+2^2}}}_{s\to s-(-4)}\\ =\frac{c}{2}{e}^{-4t}\sin 2t\end{array}$

So the y’ value$\frac{C}{2}$

$\begin{array}{c}y\text{'}\left(t\right)=\frac{c}{2}-4e^{-4t}\sin 2t+2e^{-4t}\cos 2t\\ y\text{'}\left(\pi \right)=\frac{c}{2}0+2e^{-4\pi }\\ y\text{'}\left(\pi \right)=c{e}^{-4\pi }\\ 0=c{e}^{-4\pi }\\ c=0\end{array}$

SO, $y\left(t\right)=0$

Hence,

data-custom-editor="chemistry" $y\left(t\right)=0$