Question

Question: In problems31 and 32use the Laplace transform And the procedure outlined in exampleto solve the given boundary-value problem.

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

Short answer

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

$$\begin{gathered} y\text{'}\text{'}+2y+y=0..y\text{'}\left(0\right)=2,y\left(1\right)=2 \\ L\left\{y\text{'}\text{'}\right\}=s^{2}Y-sy\left(0\right)-y\text{'}\left(0\right) \\ \begin{array}{l}{s}^{2}y-\left(sk\right)-2+2sy-2k+y=0\\ \to \frac{sk}{{\left(s+1\right)}^2}+\frac{2}{{\left(S+1\right)}^2}+\frac{2k}{{\left(s+1\right)}^2}\end{array} \end{gathered}$$

$$\begin{gathered} \begin{array}{c}y\left(t\right)=L^{-1}\frac{K}{\left(s+1\right)}+\frac{2}{{\left(s+1\right)}^2}+\frac{K}{{\left(s+1\right)}^2}\\ y\left(t\right)=K{e}^{-t}+2t{e}^{-t}+kt{e}^{-t}\end{array} \\ y\left(t\right)=e^{-t+1}(1+t)-e^{-t}(1-2t) \end{gathered}$$

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:

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

Derivatives transformation:

role="math" localid="1664303331422" $$\begin{gathered} L\left\{y\text{'}\text{'}\right\}=s^{2}Y-sy\left(0\right)-y\text{'}\left(0\right) \\ L\left\{y\text{'}\right\}=sy-y\left(0\right) \\ L\left\{y\text{'}\text{'}+2y\text{'}+y\right\}=L\left\{0\right\} \end{gathered}$$

y(0) = k then after transforming we have to solve for k.

$\begin{array}{l}{s}^{2}y-\left(sk\right)-2+2sy-2k+y=0\\ \to \frac{sk}{{\left(s+1\right)}^2}+\frac{2}{{\left(S+1\right)}^2}+\frac{2k}{{\left(s+1\right)}^2}\end{array}$

Step 2:  fraction's numerator

The part of a fraction above the line that represents the number to be divided by the denominator.

We must now convert s in the first fraction to s+1 in order to use operational properties.

I'll take one k from the last fraction's numerator and put it in the first fraction's numerator:

$$\begin{gathered} Y=\frac{sk}{{\left(s+1\right)}^2}+\frac{2}{{\left(s+1\right)}^2}+\frac{2k}{{\left(S+1\right)}^2} \\ Y=\frac{k}{\left(s+1\right)}+\frac{2}{{\left(s+1\right)}^2}+\frac{K}{{\left(s+1\right)}^2} \end{gathered}$$

Now we apply the inverse transform, remembering that multiplying by exponential in the time domain results in a shift in the s domain:

$\begin{array}{l}y\left(t\right)=L^{-1}\frac{K}{\left(s+1\right)}+\frac{2}{{\left(s+1\right)}^2}+\frac{K}{{\left(s+1\right)}^2}\\ y\left(t\right)=K{e}^{-t}+2t{e}^{-t}+kt{e}^{-t}\end{array}$

Now we need to solve for y(1) = 2 to find k.

$$\begin{gathered} \begin{array}{l}2=K{e}^{-t}+2e^{-t}+k{e}^{-t}\to k=e-1\\ y\left(t\right)=(e-1)e^{-t}+2t{e}^{-t}+(e-1)t{e}^{-t}\end{array} \\ y\left(t\right)=e^{-t+1}(1+t)-e^{-t}(1-2t) \end{gathered}$$

Hence,

$y\left(t\right)=e^{-t+1}(1+t)-e^{-t}(1-2t)$