Question

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

Equation.

38. $\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{W}\mathbf{h}\mathbf{e}\mathbf{r}\mathbf{e}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}{t}^2,0⩽t<1\\ 0,t⩾1,\end{array}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

$y\left(t\right)=-\frac{2}{125}-\frac{2}{25}t-\frac{1}{5}{t}^2+\frac{127}{125}{e}^{5t}$

$+\frac{37}{125}U(t-1)+\frac{12}{25}(t-1)U(t-1)+\frac{1}{5}(t-1{)}^{2}U(t-1)-\frac{37}{125}{e}^{5(t-1)}U(t-1)$

Step-by-step solution

Given Information

$y\text{'}-5y=f\left(t\right),Wheref\left(t\right)=\left\{\begin{array}{l}{t}^2,0⩽t<1\\ 0,t⩾1,\end{array}y\right(0)=1$

Apply Laplace Transform

The function of the type as

$f\left(t\right)=\left\{\begin{array}{l}g\left(t\right),0⩽t<a\\ h\left(t\right),t⩾a\end{array}\right.$

$f\left(t\right)=g\left(t\right)-g\left(t\right)U(t-a)+h\left(t\right)U(t-a)$

$f\left(t\right)=t^2-t^{2}U(t-1)$

$y\text{'}-5y=t^2-t^{2}U(t-1)$

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

Take Laplace transform of both side of the given equation

$$\begin{gathered} L\left\{t^2-t^{2}U(t-1)\right\}=L\left\{y\text{'}-5y\right\} \\ =L\left\{y\text{'}\right\}-5L\left\{y\right\} \end{gathered}$$

$\begin{array}{rcl}sL\left\{y\right\}-y\left(0\right)-5L\left\{y\right\}\left(y\right(0)& =& 1)\\ & =& (s-5)L\left\{y\right\}-1\end{array}$

$L\left\{t^2-t^{2}U(t-1)\right\}=(s-5)Y\left(s\right)-1$

$L\left\{t^2-t^{2}U(t-1)\right\}=L\left\{t^2\right\}-L\left\{t^{2}U(t-1)\right\}$

$=L\left\{t^2\right\}-L\left\{{(t-1)}^{2}U(t-1)\right\}+L\left\{\right[2t-2+1\left]U\right(t-1\left)\right\}$

$L\left\{t^2\right\}-L\left\{{(t-1)}^{2}U(t-1)\right\}+2L\left\{\right[t-1\left]U\right(t-1\left)\right\}-L\left\{U\right(t-1\left)\right\}$

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

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

$\left(s^2-5\right)Y\left(s\right)-1=\frac{2}{s^3}-\frac{2}{s^3}{e}^{-s}-\frac{2}{s^2}{e}^{-s}-\frac{e^{-3}}{s}$

$\left(s^2-5\right)Y\left(s\right)-1=\frac{2}{s^3}-e^{-s}\left[\frac{2+2s+s^2}{s^3}\right]$

$Y\left(s\right)=\frac{2+s^3}{s^3\left(s^2-5\right)}-e^{-s}\left[\frac{2+2s+s^2}{s^3\left(s^2-5\right)}\right]$

Apply Partial Fraction

$\frac{2+s^3}{s^3\left(s^2-5\right)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s^3}+\frac{D}{s-5}$

$\frac{A{s}^2(s-5)+Bs(s-5)+C(s-5)+D{s}^3}{s^3\left(s^2-5\right)}$

$2+s^3=A{s}^3-5A{s}^2+B{s}^2-5Bs+Cs-5C+D{s}^3$

$2+s^3=(A+D)s^3+(B-5A)s^2+(C-5B)s-5C$

$A+D=1...\left(1\right)$

$B-5A=0...\left(2\right)$

$C-5B=0...\left(3\right)$

$-5C=2...\left(4\right)$

$D=1-A$

$$\begin{gathered} A=\frac{B}{5} \\ B=\frac{C}{5} \\ C=-\frac{2}{5} \\ D=\frac{127}{125} \end{gathered}$$

$$\begin{gathered} A=-\frac{2}{125} \\ B=-\frac{2}{25} \\ C=-\frac{2}{5} \end{gathered}$$

$\frac{2+s^3}{s^3\left(s^2-5\right)}=-\frac{2}{125s}-\frac{2}{25s^2}-\frac{2}{5s^3}+\frac{127}{125(s-5)}$

$\frac{2+2s+s^2}{s^3\left(s^2-5\right)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s^3}+\frac{D}{s-5}$

$\frac{A{s}^2(s-5)+Bs(s-5)+C(s-5)+D{s}^3}{s^3\left(s^2-5\right)}$

$$\begin{gathered} 2+2s+s^2=A{s}^3-5A{s}^2+B{s}^2-5Bs+Cs-5C+D{s}^3 \\ 2+2s+s^2=(A+D)s^3+(B-5A)s^2+(C-5B)s-5C \end{gathered}$$

$\begin{array}{rcl}A+D& =& 0\\ B-5A& =& 1\\ C-5B& =& 2\\ -5C& =& 2\end{array}$

$D=-A$

$A=\frac{B-1}{5}$

$B=\frac{C-2}{5}$

$C=-\frac{2}{5}$

$D=\frac{37}{125}$

$A=-\frac{37}{125}$

$B=-\frac{12}{25}$

$C=-\frac{2}{5}$

$\frac{2+2s+s^2}{s^3\left(s^2-5\right)}=-\frac{37}{125s}-\frac{12}{25s^2}-\frac{2}{5s^3}+\frac{37}{125(s-5)}$

$Y\left(s\right)=-\frac{2}{125s}-\frac{2}{25s^2}-\frac{2}{5s^3}+\frac{127}{125(s-5)}-e^{-s}\left[-\frac{37}{125s}-\frac{12}{25s^2}-\frac{2}{5s^3}+\frac{37}{125(s-5)}\right]$

Apply Inverse Laplace Transform

$Y\left(s\right)=-\frac{2}{125s}-\frac{2}{25s^2}-\frac{2}{5s^3}+\frac{127}{125(s-5)}-e^{-s}\left[-\frac{37}{125s}-\frac{12}{25s^2}-\frac{2}{5s^3}+\frac{37}{125(s-5)}\right]$

$L^{-1}\left\{Y\right(s\left)\right\}=-\frac{2}{125}{\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}-\frac{2}{25}{L}^{-1}\left\{\frac{1}{s^2}\right\}-\frac{2}{5}{L}^{-1}\left\{\frac{1}{s^3}\right\}+\frac{127}{125}{L}^{-1}\left\{\frac{1}{s-5}\right\}+\frac{37}{125}{L}^{-1}\left\{\frac{e^{-s}}{s}\right\}$

$+\frac{12}{25}{L}^{-1}\left\{\frac{e^{-s}}{s^2}\right\}+\frac{2}{5}{L}^{-1}\left\{\frac{e^{-s}}{s^3}\right\}-\frac{37}{125}{L}^{-1}\left\{\frac{e^{-s}}{s-5}\right\}$

$\left(L^{-1}\left\{\frac{1}{s}\right\}=1,L^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}\right)$

$y\left(t\right)=-\frac{2}{125}-\frac{2}{25}t-\frac{1}{5}{t}^2+\frac{127}{125}{e}^{5t}$

$+\frac{37}{125}U(t-1)+\frac{12}{25}(t-1)U(t-1)+\frac{1}{5}(t-1{)}^{2}U(t-1)-\frac{37}{125}{e}^{5(t-1)}U(t-1)$