Question

Reread (i) in the Remarks at the end of this section. Then use

the Laplace transform to solve the initial-value problem:

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{t}}\mathit{\delta }\mathbf{(}\mathit{t}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{.}$

Use a graphing utility to graph y(t) for $\mathbf{0}\mathbf{⩽}\mathit{t}\mathbf{⩽}\mathbf{5}\mathbf{.}$

Short answer

The solution is,

$y\left(t\right)=\frac{e}{2}(e^{-(t-1)}-e^{-2(t-1)})U(t-1)-e^{-3t}+e^{-t}$

$y\left(t\right)=\frac{e}{2}(e^{-(t-1)}-e^{-2(t-1)})U(t-1)-e^{-3t}+e^{-t}$

Step-by-step solution

Step 1:Definition of Laplace Transform

Let $f$be a function defined for$t⩾0$. Then the integral

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^x{e}^{-xt}f\left(t\right)dt$

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

Step 2:Use Laplace Transform

We know that$\mathcal{L}\left\{x^{\text{'}\text{'}}\right\}=s^2\mathcal{L}\left\{x\right\}-sx\left(0\right)-x^{\text{'}}\left(0\right)$

$\mathcal{L}\left\{x^{\text{'}}\right\}=s\mathcal{L}\left\{x\right\}-x\left(0\right)$

$$" width="9">Ly'=sL{y}-y(0)

Take the Laplace transform on both sides of each given differential equation

$\mathcal{L}\left\{e^t\delta (t-1)\right\}=\mathcal{L}\left\{y^{\text{'}\text{'}}\right\}+4\mathcal{L}\left\{y\text{'}\right\}+3\mathcal{L}\left\{y\right\}$

Substitute the known values,

$\mathcal{L}\left\{e^t\delta (t-1)\right\}=s^2\mathcal{L}\left\{y\right\}-sy\left(0\right)-y^{\text{'}}\left(0\right)+4(s\mathcal{L}\{y\}-y(0\left)\right)+3\mathcal{L}\left\{y\right\}$

$\mathcal{L}\left\{e^t\delta (t-1)\right\}=(s^2+4s+3)\mathcal{L}\left\{y\right\}-2$$\left(y\left(0\right)=0,y^{\text{'}}\left(0\right)=2\right)$

Denote$\mathcal{L}\left\{y\right\}=Y\left(s\right)$in the above system,

$\mathcal{L}\left\{e^t\delta (t-1)\right\}=(s^2+4s+3)Y\left(s\right)-2$

width="244">Letδ(t-1)=(s2+4s+3)Y(s)-2......(1)

unit function.

The formula for Laplace transform is,

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

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

width="146">L{f(t)}=∫0∞e-stf(t)dt

We have that $f\left(t\right)=e^t\delta (t-1)$. So using the above formula, we get

$L\left\{e^t\delta \right(t-1\left)\right\}=\underset{0}{\overset{\infty }{\int }}{e}^{-st}{e}^t\delta (t-1)dt$

We can write as:

$L\left\{e^t\delta \right(t-1\left)\right\}=\underset{-\infty }{\overset{\infty }{\int }}{e}^{-st}U\left(t\right)e^t\delta (t-1)dt$

Where $u\left(t\right)$, is the unit function defined as,

$$\begin{gathered} U\left(t\right)=\left\{0,t<0 \\ 1,t⩾0 \\ \right. \end{gathered}$$

width="106">U(t)=0,t<01,t⩾0

Using the formula,

$\underset{-\infty }{\overset{\infty }{\int }}\delta (t-t_0)f\left(t\right)dt=f\left(t_0\right)$

We get,

$$\begin{gathered} L\left\{e^t\delta \right(t-1\left)\right\}=\underset{-\infty }{\overset{\infty }{\int }}\delta (t-1)e^{-st}U\left(t\right)e^{t}dt \\ =U\left(1\right)e^{-s}e \\ =e^{1-s} \end{gathered}$$

So, the equation (1) becomes,

$$\begin{gathered} (s^2+4s+3)Y\left(s\right)-2=e^{1-s} \\ Y\left(s\right)=e^{1-s}\frac{1}{(s^2+4s+3)}+\frac{2}{(s^2+4s+3)} \end{gathered}$$

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

Step 4:use the partial fraction technique.

Decompose the fraction$\frac{1}{(s+3)(s+1)}$using the partial fraction technique.

$$\begin{gathered} \frac{1}{(s+3)(s+1)}=\frac{A}{(s+3)}+\frac{B}{(s+1)}.........\left(3\right) \\ =\frac{A(s+1)+B(s+3)}{(s+3)(s+1)} \end{gathered}$$

Since the fraction in above equation have the same denominators, it takes their numerators to be equal.

$1=A(s+1)+B(s+3)$

If, $s=-1$, then

$$\begin{gathered} 1=2B \\ B=\frac{1}{2} \end{gathered}$$

If, $s=-3$then,

$$\begin{gathered} 1=-2A \\ A=-\frac{1}{2} \end{gathered}$$

Substitute A and B in the equation (3), we get

$\frac{1}{(s+3)(s+1)}=-\frac{1}{2}\frac{1}{(s+3)}+\frac{1}{2}\frac{1}{(s+1)}$

Then the equation 2 becomes,

$Y\left(s\right)=e^{1-s}\left[-\frac{1}{2}\frac{1}{(s+3)}+\frac{1}{2}\frac{1}{(s+1)}\right]+2\left[-\frac{1}{2}\frac{1}{(s+3)}+\frac{1}{2}\frac{1}{(s+1)}\right]$

$Y\left(s\right)=\left[-\frac{e}{2}\frac{e^{-s}}{(s+3)}+\frac{e}{2}\frac{e^{-s}}{(s+1)}\right]-\frac{1}{(s+3)}+\frac{1}{(s+1)}$

Step 5:Apply inverse Laplace transform.

Apply inverse Laplace transform to each side.

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{Y\right(s\left)\right\}={\mathcal{L}}^{-1}\left\{\left[-\frac{e}{2}\frac{e^{-s}}{(s+3)}+\frac{e}{2}\frac{e^{-s}}{(s+1)}\right]-\frac{1}{(s+3)}+\frac{1}{(s+1)}\right\} \\ y\left(t\right)=\frac{e}{2}(e^{-(t-1)}-e^{-2(t-1)})U(t-1)-e^{-3t}+e^{-t} \end{gathered}$$

Therefore,

localid="1663905033596" $y\left(t\right)=\frac{e}{2}(e^{-(t-1)}-e^{-2(t-1)})U(t-1)-e^{-3t}+e^{-t}$

The graph of localid="1663905048072" $y\left(t\right)$is shown below,

Hence, the solution is Hence, the solution is$y\left(t\right)=\frac{e}{2}(e^{-(t-1)}-e^{-2(t-1)})U(t-1)-e^{-3t}+e^{-t}$