Question

In Problems 63-70 use the Laplace transform to solve the given initial-value problem.

$y\text{'}\text{'}-5y\text{'}+6y=\mathcal{U}(t-1),y\left(0\right)=0,y\text{'}\left(0\right)=1$

Short answer

The solution for the initial-value problem$y\text{'}\text{'}-5y\text{'}+6y=\mathcal{U}(t-1),y\left(0\right)=0,y\text{'}\left(0\right)=1$using Laplace transform is$y\left(t\right)=\left(\frac{1}{6}-\frac{1}{2}{e}^{2(t-1)}+\frac{1}{3}{e}^{3(t-1)}\right)U(t-1)-e^{2t}+e^{3t}$ .

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. The Laplace transform$\mathcal{L}$of a function$f\left(t\right)$is defined as

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

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

Find the Laplace transform for the given expression.

Use Laplace transform on both sides of the expression.

$$\begin{gathered} L\left\{y\text{'}\text{'}-5y\text{'}+6y\right\}=L\left\{U\right(t-1\left)\right\}\dots \dots y\left(0\right)=0,y\text{'}\left(0\right)=1 \\ {s}^{2}Y\left(s\right)-s\left(0\right)-\left(1\right)+5\left(sY\right(s)-0)+6Y\left(s\right)=\frac{e^{-s}}{s} \\ Y\left(s\right)\left(s^2-5s+6\right)=\frac{e^{-s}}{s}+1 \\ Y\left(s\right)=\frac{e^{-s}}{s\left(s^2-5s+6\right)}+\frac{1}{s^2-5s+6} \end{gathered}$$

The Laplace transform $y\text{'}\text{'}-5y\text{'}+6y=\mathcal{U}(t-1),y\left(0\right)=0,y\text{'}\left(0\right)=1$before decomposition is$Y\left(s\right)=\frac{e^{-s}}{s\left(s^2-5s+6\right)}+\frac{1}{s^2-5s+6}$ .

Find the decomposed value for the given expression.

Decomposing,

$$\begin{gathered} Y\left(s\right)=\frac{e^{-s}}{s\left(s^2-5s+6\right)}+\frac{1}{s^2-5s+6} \\ \frac{1}{s\left(s^2-5s+6\right)}=\frac{a}{s}+\frac{b}{s-2}+\frac{c}{s-3}=\frac{1}{6s}-\frac{1}{2(s-2)}+\frac{1}{3(s-3)} \\ \frac{1}{\left(s^2-5s+6\right)}=\frac{a}{s-2}+\frac{b}{s-3}=\frac{-1}{s-2}+\frac{1}{s-3} \\ Y\left(s\right)=\left(\frac{1}{6s}-\frac{1}{2(s-2)}+\frac{1}{3(s-3)}\right)e^{-s}+\frac{-1}{s-2}+\frac{1}{s-3} \end{gathered}$$

Hence, the decomposed value of $Y\left(s\right)=\frac{e^{-s}}{s\left(s^2-5s+6\right)}+\frac{1}{s^2-5s+6}$is $Y\left(s\right)=\left(\frac{1}{6s}-\frac{1}{2(s-2)}+\frac{1}{3(s-3)}\right)e^{-s}+\frac{-1}{s-2}+\frac{1}{s-3}$

Find the inverse Laplace value of the given expression.

Apply inverse Laplace transform.

$$\begin{gathered} L^{-1}\left\{F\left(s\right)e^{-as}\right\}=f(t-a)U(t-a) \\ y\left(t\right)=\left(\frac{1}{6}-\frac{1}{2}{e}^{2(t-1)}+\frac{1}{3}{e}^{3(t-1)}\right)U(t-1)-e^{2t}+e^{3t} \end{gathered}$$

The inverse Laplace transform of

role="math" localid="1663929161436" $$\begin{gathered} Y\left(s\right)=\left(\frac{1}{6s}-\frac{1}{2(s-2)}+\frac{1}{3(s-3)}\right)e^{-s}+\frac{-1}{s-2}+\frac{1}{s-3}is \\ y\left(t\right)=\left(\frac{1}{6}-\frac{1}{2}{e}^{2(t-1)}+\frac{1}{3}{e}^{3(t-1)}\right)U(t-1)-e^{2t}+e^{3t} \end{gathered}$$