Question

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

$y\text{'}+2y=f\left(t\right),y\left(0\right)=0,wheref\left(t\right)=f\left(t\right)=\{t,0\le t<10,t\ge 1$

Short answer

The Laplace transform of the given function is$y\left(t\right)=-1/4+1/2t+1/4e^{(-2t)}-1/4U(t-1)-1/2(t-1)U(t-1)+1/4e^{(-2(}t-1\left)\right)U(t-1).$

Step-by-step solution

Definition of Laplace Transform

Given that,

$L\left\{y\text{'}+2\text{y}\right\}=L\left\{f\left(t\right)\right\}\dots \dots y\left(0\right)=0.$

Where,

$$\begin{gathered} f\left(t\right)=\left\{\begin{array}{ll}0,& 0⩽t<1\\ 5,& t⩾1\end{array}\right. \\ f\left(t\right)=t-tU(t-1) \end{gathered}$$

First, we need to replace “t” that is multiplied by the step function with $\left(t-1\right)$

$f\left(t\right)=t-\left(\left(t-1\right)+1\right)U\left(t-1\right)=t-\left(t-1\right)U\left(t-1\right)-U\left(t-1\right).$

$$\begin{gathered} sY-0+2Y=\frac{1}{s^2}-\frac{e^{-s}}{s^2}-\frac{e^{-s}}{s} \\ Y=\frac{1}{s^2(s+2)}-\frac{e^{-s}}{s^2(s+2)}-\frac{e^{-s}}{s(s+2)} \end{gathered}$$

1. Decomposing

$$\begin{gathered} \frac{1}{s^2(s+2)}=\frac{a}{s}+\frac{b}{s^2}+\frac{c}{(s+2)} \\ =-\frac{1}{4s}+\frac{1}{2s^2}+\frac{1}{4(s+2)} \end{gathered}$$

2. Decomposing

$$\begin{gathered} \frac{1}{S(S+2)}=\frac{a}{s}+\frac{b}{(S+2)}=\frac{1}{2s}-\frac{1}{2(S+2)} \\ Y=-\frac{1}{4s}+\frac{1}{2s^2}+\frac{1}{4(s+2)}+\frac{e^{-s}}{4s}-\frac{e^{-s}}{2s^2}-\frac{e^{-s}}{4(s+2)}-\frac{e^{-s}}{2s}-\frac{e^{-s}}{2(s+2)} \end{gathered}$$

Find the value of Y

Given that,

$\mathrm{L}\{\mathrm{y}\text{'}+2\text{y}\}=\mathrm{L}\left\{\mathrm{f}\left(\mathrm{t}\right)\right\}\dots \dots \mathrm{y}\left(0\right)=0.$

Where,

First, we need to replace “t” that is multiplied by the step function with $(t-1)$

$f\left(t\right)=t-((t-1)+1)U(t-1)=t-(t-1)U(t-1)-U(t-1).$

1. Decomposing

2. Decomposing

$\frac{1}{\mathrm{S}(\mathrm{S}+2)}=\frac{\mathrm{a}}{\mathrm{s}}+\frac{\mathrm{b}}{(\mathrm{S}+2)}=\frac{1}{2\mathrm{s}}-\frac{1}{2(\mathrm{S}+2)}$

$\mathrm{Y}=-\frac{1}{4\mathrm{s}}+\frac{1}{2{\mathrm{s}}^2}+\frac{1}{4(\mathrm{s}+2)}+\frac{{\mathrm{e}}^{-\mathrm{s}}}{4\mathrm{s}}-\frac{{\mathrm{e}}^{-\mathrm{s}}}{2{\mathrm{s}}^2}-\frac{{\mathrm{e}}^{-\mathrm{s}}}{4(\mathrm{s}+2)}-\frac{{\mathrm{e}}^{-\mathrm{s}}}{2\mathrm{s}}-\frac{{\mathrm{e}}^{-\mathrm{s}}}{2(\mathrm{s}+2)}$

Unit Step Function

The unit step function $u(t-a)$is defined to b

$$\begin{gathered} u\left(t-a\right)=\left\{\begin{array}{ll}0,& 0⩽t<1\\ 1,& t⩾1\end{array}\right. \\ -\frac{1}{4s}+\frac{1}{2s^2}+\frac{1}{4(s+2)}+\frac{e^{-s}}{4s}-\frac{e^{-s}}{2s^2}-\frac{e^{-s}}{4(s+2)}-\frac{e^{-s}}{2s}-\frac{e^{-s}}{2(s+2)} \\ y\left(t\right)=-\frac{1}{4}+\frac{1}{2}t+\frac{1}{4}{e}^{-2t}+\frac{1}{4}U(t-1)-\frac{1}{2}(t-1)U(t-1)-\frac{1}{4}{e}^{-2(t-1)}U(t-1) \end{gathered}$$