Question

In Problems 35–42 use the Laplace transform to solve the given

equation.

39.$\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}$$\mathbf{w}\mathbf{h}\mathbf{e}\mathbf{r}\mathbf{e}\mathbf{}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{}\mathbf{i}\mathbf{s}\mathbf{g}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{i}\mathbf{n}\mathbf{F}\mathbf{i}\mathbf{g}\mathbf{u}\mathbf{r}\mathbf{e}\mathbf{7}\mathbf{.}\mathbf{R}\mathbf{.}\mathbf{10}\mathbf{.}\mathbf{}$

Short answer

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

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

Step-by-step solution

Given Information

$y\text{'}+2y=f\left(t\right),y\left(0\right)=1$

Apply Laplace Transform

The function of the type as

$f\left(t\right)=\left\{\begin{array}{c}0,0\le t<a\\ g\left(t\right),a\le t<b\\ h\left(t\right),b\le t<c\\ 0,t\ge c\end{array}\right.$

$f\left(t\right)=g\left(t\right)\left[U\right(t-a)-U(t-b\left)\right]+h\left(t\right)\left[U\right(t-b)-U(t-c\left)\right]$

The given function is

$f\left(t\right)=\left\{\begin{array}{c}0,0\le t<1\\ t-1,1\le t<2\\ 2-(t-1),2\le t<3\\ 0,t\ge 3\end{array}\right.$

$f\left(t\right)=(t-1)\left[U\right(t-1)-U(t-2\left)\right]+(2-(t-1\left)\right)\left[U\right(t-2)-U(t-3\left)\right]$

$y\text{'}+2y=(t-1)U(t-1)-(2t-4)U(t-2)+(t-3)U(t-3)$

Take Laplace transform of both side of the given equation

$L\left\{y\text{'}\right\}=sL\left\{y\right\}-y\left(0\right)$

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

$$\begin{gathered} sL\left\{y\right\}-y\left(0\right)+2L\left\{y\right\}\left(y\right(0)=1) \\ =(s+2)L\left\{y\right\}-1 \end{gathered}$$

$L\left\{\right(t-1\left)U\right(t-1)-(2t-4\left)U\right(t-2)+(t-3\left)U\right(t-3\left)\right\}=(s+2)Y\left(s\right)-1$

$L\left\{\right(t-1\left)U\right(t-1)-(2t-4\left)U\right(t-2)+(t-3\left)U\right(t-3\left)\right\}$

$L\left\{\right(t-1\left)U\right(t-1)-(2t-4\left)U\right(t-2)+(t-3\left)U\right(t-3\left)\right\}=L\left\{\right(t-1\left)U\right(t-1\left)\right\}-2L\left\{\right(t-2\left)U\right(t-2\left)\right\}+L(t-3)U(t-3)\}$

$\frac{1}{s^2}{e}^{-s}-\frac{2}{s^2}{e}^{-2s}+\frac{1}{s^2}{e}^{-3s}$

$\left\{\begin{array}{l}\text{If}\mathrm{F}\left(\mathrm{s}\right)=\mathrm{L}\left\{\mathrm{f}\right(\mathrm{t}\left)\right\}\text{and}\mathrm{a}>0,\text{thenL}\left\{\mathrm{f}\right(\mathrm{t}-\mathrm{a}\left)\mathrm{U}\right(\mathrm{t}-\mathrm{a}\left)\right\}={\mathrm{e}}^{-\mathrm{as}}\mathrm{F}\left(\mathrm{s}\right)\\ \mathrm{L}\left\{\mathrm{U}\right(\mathrm{t}-\mathrm{a}\left)\right\}=\frac{{\mathrm{e}}^{-\mathrm{as}}}{\mathrm{s}}\\ \mathrm{L}\left\{{\mathrm{t}}^{\mathrm{n}}\right\}=\frac{\mathrm{n}!}{{\mathrm{s}}^{\mathrm{n}+1}}\end{array}\right.$

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

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

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

Apply Partial Fraction

$\frac{1}{s(s+2)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s+2}$

$=\frac{As(s+2)+B(s+2)+C{s}^2}{s^2(s+2)}$

$1=A{s}^2+2As+Bs+2B+C{s}^2$

$1=(A+C)s^2+(2A+B)s+2B$

$A+C=0..\left(1\right)$

$2A+B=0...\left(2\right)$

$2B=1.....\left(3\right)$

From the first equation

$C=-A$

From the second equation as

$B=\frac{1}{2}$

From the third equation as

$A=-\frac{B}{2}$

Substitute these values are $C=\frac{1}{4},A=-\frac{1}{4},B=\frac{1}{2}$

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

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

Apply Inverse Laplace Transform

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

$e^{-2s}+\left[-\frac{1}{4}{L}^{-1}\left\{\frac{1}{s}\right\}+\frac{1}{2}{L}^{-1}\left\{\frac{1}{s^2}\right\}+\frac{1}{4}{L}^{-1}\left\{\frac{1}{s+2}\right\}\right]e^{-3s}+L^{-1}\left\{\frac{1}{s+2}\right\}$

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

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