Question

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

equation.

40. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{w}\mathbf{h}\mathbf{e}\mathbf{r}\mathbf{e}$

$\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{12}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{x}}{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{k}}\mathbf{\times }\mathbf{t}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{k}\mathbf{)}\mathbf{.}$

Short answer

$y\left(t\right)=3E\left(t\right)-3e^{-t}-4e^{-t}\sum _{k-1}^{\infty }{(-1)}^{k}U(t-k)+{e^{-u}}_{k-1}^{\infty }(-1{)}^{k}U(t-k)$

Step-by-step solution

To Find the laplace transform of both side.

$\begin{array}{rcl}(*)L\left\{y\text{'}\text{'}\right\}& =& s^{2}L\left\{y\right\}-sy\left(0\right)-y^{\text{'}}\left(0\right)\\ L\left\{y\text{'}\right\}& =& sL\left\{y\right\}-y\left(0\right)\end{array}$

$LetE\left(t\right)=\sum _{k=0}^{\infty }{(-1)}^{k}U(t-k).$

Take Laplace transform of both side of the given equation.

$*12L\left\{E\right(t\left)\right\}=L\left\{y\text{'}\text{'}+5y\text{'}+4y\right\}$

$=L\left\{y\text{'}\text{'}+5L\{y\text{'}\}+4L\{y\right\}$

$$\begin{gathered} =s^{2}C\left\{y\right\}-sy\left(0\right)-y^{\text{'}}\left(0\right)+5sC\left\{y\right\}-5y\left(0\right)+4C\left\{y\right\} \\ =\left(s^2+5s+4\right)L\left\{y\right\}-3\left(y\left(0\right)=0,y^{\text{'}}\left(0\right)=3\right) \end{gathered}$$

Denote $L\left\{y\right\}=Y\left(s\right)$ and replace in the above equation.

$12L\left\{E\right(t\left)\right\}=\left(s^2+5s+4\right)Y\left(s\right)-3$ . . . (i)

$C\left\{E\right(t\left)\right\}$ can be written as:

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

$=L\left\{U\right(t-0\left)\right\}-L\left\{U\right(t-1\left)\right\}+C\left\{U\right(t-2\left)\right\}-L\left\{U\right(t-3\left)\right\}+.....$

$=\frac{1}{s}-\frac{e^{--}}{s}+\frac{e^{-2}}{s}-\frac{e^{-3}}{s}+L\left\{U\right(t-a\left)\right\}=\frac{e^{-as}}{s}$

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

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

Now substitute the above equation (1)

$12\frac{1}{s\left(1+e^{-s}\right)}=\left(s^2+5s+4\right)Y\left(s\right)-3$

$\left(s^2+5s+4\right)Y\left(s\right)=\frac{12}{s\left(1+e^{-8}\right)}+3$

$(s+1)(s+4)Y\left(s\right)=\frac{12}{s\left(1+e^{-s}\right)}+3$

$Y\left(s\right)=\frac{12}{\left(1+e^{-s}\right)}\frac{1}{s(s+1)(s+4)}+\frac{3}{(s+1)(s+4)}$

Find the partial fraction technique.

Decompose the fraction $\frac{1}{s(s+1)(s+4)}$ using the partial fraction technique.

$\frac{1}{s(s+1)(s+4)}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{s+4}$ ……… (3)

$=\frac{A(s+1)(s+4)+Bs(s+4)+Cs(s+1)}{s(s+1)(s+4)}$

Since the fractions in above equation have the same denominators, it takes their numerators to be equal.

$A\left(s+1\right)\left(s+4\right)+Bs\left(s+4\right)+Cs\left(s+1\right)$

$$\begin{gathered} Ifs=-1,then \\ 1=-3B \\ B=-\frac{{\displaystyle 1}}{{\displaystyle 3}} \end{gathered}$$

$$\begin{gathered} ifs=4,then \\ 1=12C \\ C=\frac{{\displaystyle 1}}{{\displaystyle 12}} \end{gathered}$$

$$\begin{gathered} ifs=0,then \\ 1=4A \\ A=\frac{{\displaystyle 1}}{{\displaystyle 4}} \end{gathered}$$

Substituting A, B and C in the equation (3), we get

$\frac{1}{s(s+1)(s+4)}=\frac{1}{4}\frac{1}{s}-\frac{1}{3}\frac{1}{s+1}+\frac{1}{12}\frac{1}{s+4}$

Decompose the fraction $\frac{3}{(s+1)(s+4)}$ using the partial fraction techrique.

$$\begin{gathered} =\frac{3}{(s+1)(s+4)}=\frac{A}{s+1}+\frac{B}{s+4} \\ =\frac{A(s+4)+B(s+1)}{(s+1)(s+4)} \end{gathered}$$

Since the fractions in above equation have the same denominators, it takes their numerators to be equal.

$3=A\left(s+4\right)+B\left(s+1\right)$

$$\begin{gathered} Ifs=-1,then \\ 3=3A \\ A=1 \end{gathered}$$

If s = -4, then

3=-3B

B=-1

Substituting A and B in the equation (4), we get

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

Step 3:Final proof

The equation (2) becomes.

$Y\left(s\right)=\frac{12}{\left(1+e^{-x}\right)}\frac{1}{4}\frac{1}{s}-\frac{1}{3}\frac{1}{s+1}+\frac{1}{12}\frac{1}{s+4}+\frac{1}{s+1}-\frac{1}{s+4}$

$Y\left(s\right)=\frac{12}{\left(1+e^{-x}\right)}\frac{1}{4}\frac{1}{s}-\frac{1}{3}\frac{1}{s+1}+\frac{1}{12}\frac{1}{s+4}+\frac{1}{s+1}-\frac{1}{s+4}$

$=\frac{3}{s\left(1+e^{-x}\right)}-\frac{4}{s+1}\frac{1}{\left(1+e^{-x}\right)}+\frac{1}{s+4}\frac{1}{\left(1+e^{-x}\right)}+\frac{1}{s+1}-\frac{1}{s+4}$

$=\frac{3}{s\left(1+e^{-x}\right)}-\frac{4}{s+1}1-e^{-s}+e^{-2s}-e^{-st}+.....$

$+\frac{1}{s+4}1-e^{-s}+e^{-2s}-e^{-2s}+........+L+\frac{1}{s+1}-\frac{1}{s+4}$

Take inverse Laplace transform of both side.

$v\left(t\right)=3{\mathcal{L}}^{-1}\frac{1}{s\left(1+e^{-s}\right)}-4{\mathcal{L}}^{-1}\frac{1}{s+1}1-e^{-t}+e^{-2t}-e^{-3t}+\cdots$

$v\left(t\right)=3L^{-1}\frac{1}{s\left(1+e^{-s}\right)}-4L^{-1}\frac{1}{s+1}1-e^{-t}+e^{-2t}-e^{-3t}+.......$

$+L^{-1}\frac{1}{s+4}1-e^{-s}+e^{-2s}-e^{-3s}+L+L^{-1}\frac{1}{s+1}-L^{-1}\frac{1}{s+4}$

$\stackrel{*+1}{=}3E\left(t\right)+e^{-t}-e^{-t}-4e^{-t}-e^{-t}U(t-1)+e^{-t}U(t-2)-e^{-t}U(t-3)+.....$

$+e^{-4t}-e^{-t}U(t-1)+e^{-4t}U(t-2)-e^{-4t}U(t-3)+....$

$=3E\left(t\right)-3e^{-t}-4{e^{-t}}_{k=1}^{\infty }(-1{)}^{k}U(t-k)+e^{-2}(-1{)}^{k}U(t-k)$

$If{L}^{-1}\left\{F\right(s\left)\right\}=f\left(t\right)anda>0,thenf(t-a)U(t-a)=L\left\{e^{-1s}F\left(s\right)\right\}(e*)$

$1e^{at}=L^{-1}\frac{1}{s-a}$

So,

$y\left(t\right)=3E\left(t\right)-3e^{-t}-4{e^{-t}}_{k-1}^{\infty }(-1{)}^{k}U(t-k)+{e^{-t}}_{k-1}^{\infty }(-1{)}^{k}U(t-k)$