Question

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

$$\begin{gathered} y\text{'}\text{'}+4y\text{'}+3y=1-u(t-2)-u(t-4)+u(t-6) \\ y\left(0\right)=0,y\text{'}\left(0\right)=0 \end{gathered}$$

Short answer

The solution for the initial-value problem $y\text{'}\text{'}+4y\text{'}+3y=1-u(t-2)-u(t-4)+u(t-6)$using Laplace transform is

$y\left(t\right)=\frac{1}{3}+\frac{1}{6}{e}^{-3t}-\frac{1}{2}{e}^{-t}-\left[\left(\frac{1}{3}+\frac{1}{6}{e}^{-3(t-2)}-\frac{1}{2}{e}^{-(t-2)}\right)U(t-2)\right]-\left[\left(\frac{1}{3}+\frac{1}{6}{e}^{-3(t-4)}-\frac{1}{2}{e}^{-(t-4)}\right)U(t-4)\right]+$

$\left[\left(\frac{1}{3}+\frac{1}{6}{e}^{-3(t-6)}-\frac{1}{2}{e}^{-(t-6)}\right)U(t-6)\right]$

Step-by-step solution

Define Laplace transform.

When specific initial conditions are supplied, especially when the initial values are zero, the Laplace transform is a handy method of solving certain types of differential equations. Laplace transform $\mathcal{L}$of a function f (t) is defined as $\mathcal{L}\left\{f\right(t\left)\right\}=\int {}_0{}^{\mathrm{\mathbb{N}}}{e}^{-st}f\left(t\right)dt$

In words we can describe this expression as the Laplace transform of f(t) equals function F of S.

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

Find the unit step function for the given expression.

The unit step function is defined as

$u\left(t\right)=\left\{\begin{array}{ll}0& t<0\\ 1& t>0\end{array}\right.$

Where

u is a function of time f.

$$\begin{gathered} L\left\{y\text{'}\text{'}+4y\text{'}+3y\right\}= \\ L\{1-U(t-2)-U(t-4)+U(t-6\left)\right\}\dots \dots \dots y\left(0\right)= \\ 0,y\text{'}\left(0\right)=0 \end{gathered}$$

The unit step function of$y\text{'}\text{'}+4y\text{'}+3y=1-u(t-2)-u(t-4)+u(t-6)$

$L\{1-U(t-2)-U(t-4)+U(t-6\left)\right\}\dots \dots \dots y\left(0\right)$

Find the Laplace transform for the given expression.

Use the Laplace transform on both sides of the expression.

$s^{2}Y-s\left(0\right)-0+4(sY-0)+3Y=\frac{1}{s}-\frac{e^{-2s}}{s}-\frac{e^{-4s}}{s}+$

Factor then isolate Y.

$$\begin{gathered} Y\left(s\right)=\frac{1}{s(s+3)(s+1)}-\frac{e^{-2s}}{s(s+3)(s+1)} \\ \frac{e^{-4s}}{s(s+3)(s+1)}+\frac{e^{-6s}}{s(s+3)(s+1)} \end{gathered}$$

The value of $L\{1-U(t-2)-U(t-4)+U(t-6\left)\right\}\dots \dots \dots y\left(0\right)$before decomposing is

$$\begin{gathered} Y\left(s\right)=\frac{1}{s(s+3)(s+1)}-\frac{e^{-2s}}{s(s+3)(s+1)} \\ \frac{e^{-4s}}{s(s+3)(s+1)}+\frac{e^{-6s}}{s(s+3)(s+1)} \end{gathered}$$

Find the decomposed value of the given expression.

$$\begin{gathered} Y\left(s\right)=\frac{1}{s(s+3)(s+1)}-\frac{e^{-2s}}{s(s+3)(s+1)} \\ \frac{e^{-4s}}{s(s+3)(s+1)}+\frac{e^{-6s}}{s(s+3)(s+1)} \end{gathered}$$

Decomposing

$$\begin{gathered} \frac{1}{s(s+3)(s+1)}=\frac{a}{s}+\frac{b}{s+3}+\frac{c}{s+1}=\frac{1}{3s}+\frac{1}{6(s+3)}-\frac{1}{2(s+1)} \\ Y\left(s\right)=\left[\frac{1}{3s}+\frac{1}{6(s+3)}-\frac{1}{2(s+1)}\right]\left(1-e^{-2s}-e^{-4s}+e^{-6s}\right) \end{gathered}$$

Hence, the decomposed value of the given expression is

$Y\left(s\right)=\left[\frac{1}{3s}+\frac{1}{6(s+3)}-\frac{1}{2(s+1)}\right]\left(1-e^{-2s}-e^{-4s}+e^{-6s}\right)$

Substituting the decomposed value into the given expression to get the solution.

Substitute $Y\left(s\right)=\left[\frac{1}{3s}+\frac{1}{6(s+3)}-\frac{1}{2(s+1)}\right]\left(1-e^{-2s}-e^{-4s}+e^{-6s}\right)$in the given expression.

$$\begin{gathered} L^{-1}\left\{\left[\frac{1}{3s}+\frac{1}{6(s-(-3\left)\right)}-\frac{1}{2(s-(-1\left)\right)}\right]\left(1-e^{-2s}-e^{-4s}+e^{-6s}\right)\right\} \\ y\left(t\right)=\frac{1}{3}+\frac{1}{6}{e}^{-3t}-\frac{1}{2}{e}^{-t}-\left[\left(\frac{1}{3}+\frac{1}{6}{e}^{-3(t-2)}-\frac{1}{2}{e}^{-(t-2)}\right)U(t-2)\right]-\left[\left(\frac{1}{3}+\frac{1}{6}{e}^{-3(t-4)}-\frac{1}{2}{e}^{-(t-4)}\right)U(t-4)\right]+ \\ \left[\left(\frac{1}{3}+\frac{1}{6}{e}^{-3(t-6)}-\frac{1}{2}{e}^{-(t-6)}\right)U(t-6)\right] \end{gathered}$$

Hence, the Laplace transform for the given initial value problem is

$$\begin{gathered} y\left(t\right)=\frac{1}{3}+\frac{1}{6}{e}^{-3t}-\frac{1}{2}{e}^{-t}-\left[\left(\frac{1}{3}+\frac{1}{6}{e}^{-3(t-2)}-\frac{1}{2}{e}^{-(t-2)}\right)U(t-2)\right]-\left[\left(\frac{1}{3}+\frac{1}{6}{e}^{-3(t-4)}-\frac{1}{2}{e}^{-(t-4)}\right)U(t-4)\right]+ \\ \left[\left(\frac{1}{3}+\frac{1}{6}{e}^{-3(t-6)}-\frac{1}{2}{e}^{-(t-6)}\right)U(t-6)\right] \end{gathered}$$