Q31E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.

$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{=}\mathit{g}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{:} \\ \mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,} \\ \mathbf{ }\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{ }\mathit{g}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}0 , 0⩽t<1,\\ t , 1<t<5,\\ 1 ,5<t\end{array} \end{gathered}$$

 

Step-by-Step Solution

Answer

The solution of the given initial value problem using the method of Laplace transforms is

 $\left.y\left(t\right)=2e^{-2t}-2e^{-3t}+\left[\frac{1}{36}+\frac{1}{6}(t-1)-\frac{1}{4}{e}^{-2(t-1)}+\frac{2}{9}{e}^{-3(t-1)}\right]u(t-1)-\left[\frac{19}{36}+\frac{1}{6}(t-5)-\frac{7}{4}{e}^{-2(t-5)}+\frac{11}{9}{e}^{-3(t-5)}\right]u(t-5)\right]$

Step 1: Define Laplace Transform

The use of Laplace transformation is to convert differential equations into algebraic equations.T   

$\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{+}\mathbf{\infty }}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{·}{\mathbf{e}}^{\mathbf{-}\mathbf{s}\mathbf{·}\mathbf{t}}\mathbf{d}\mathit{t}$

Where, F(s)= Laplace transformation

S= Complex number

t= real number>=0

t'= first derivative of teh function f(t)

 

Step 2: Apply Laplace transform:

Given initial value problem

$y\text{'}\text{'}+5y\text{'}+6y=g\left(t\right)$

where Also

$$\begin{gathered} g\left(t\right)=t,1<t<5 \\ =1,5<t \end{gathered}$$

Using rectangular and unit function we can write

$$\begin{gathered} g\left(t\right)=t\left[u\right(t-1)-u(t-5\left)\right]+1u(t-5) \\ =tu(t-1)-(t-1)u(t-5) \end{gathered}$$$y\left(0\right)=0 and y\text{'}\left(0\right)=2$

Taking Laplace transform of initial value problem is

$\mathcal{L}y\text{'}\text{'}\left(s\right)+5\mathcal{L}y\text{'}\left(s\right)+6\mathcal{L}y\left(s\right)=\mathcal{L}\left[g\right(t\left)\right]$

$$\begin{gathered} s^2\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+5s\mathcal{L}y\left(s\right)-5y\left(0\right)+6y\left(s\right)=\mathcal{L}\left[tu\right(t-1)-(t-1\left)u\right(t-5\left)\right] \\ {s}^2\mathcal{L}y\left(s\right)+0-2+5s\mathcal{L}y\left(s\right)-0+6\mathcal{L}y\left(s\right)=e^{-8}\left[\frac{1}{s^2}+\frac{1}{s}\right]-e^{-5s}\left[\frac{1}{s^2}+\frac{4}{s}\right]\left(s^2+5s+6\right)\mathcal{L}y\left(s\right)-2 \\ =\left[\frac{e^{-8}(1+s)}{s^2}\right]-\left[\frac{e^{-5s}(1+4s)}{s^2}\right] \end{gathered}$$

$$\begin{gathered} \mathcal{L}y\left(s\right)=\frac{2}{\left(s^2+5s+6\right)}+\left[\frac{e^{-A}(1+s)}{s^2\left(s^2+5s+6\right)}\right]-\left[\frac{e^{-5s}(1+1s)}{s^2\left(s^2+5s+6\right)}\right] \\ =\frac{2}{s+2}-\frac{2}{s+3}+\left[\frac{e^s(1+s)}{s^2\left(s^2+5s+6\right)}\right]-\left[\frac{e^{5s}(1+4s)}{s^2\left(s^2+5s+6\right)}\right]\cdots \left(1\right) \end{gathered}$$

 

Using partial fraction

$$\begin{gathered} \frac{1+s}{s^2\left(s^2+5s+6\right)}=\frac{\frac{1}{36}}{s}+\frac{\frac{1}{6}}{s^2}-\frac{\frac{1}{4}}{s+2}+\frac{\frac{2}{9}}{s+3} \\ \frac{1+4s}{s^2\left(s^2+5s+6\right)}=\frac{\frac{19}{36}}{s}+\frac{\frac{1}{6}}{s^2}-\frac{\frac{7}{4}}{s+2}+\frac{\frac{11}{9}}{s+3} \end{gathered}$$

 

Equation first becomes as

$\mathcal{L}y\left(s\right)=\frac{2}{s+2}-\frac{2}{s+3}+e^{-s}\left[\frac{\frac{1}{36}}{s}+\frac{\frac{1}{6}}{s^2}-\frac{\frac{1}{4}}{s+2}+\frac{\frac{2}{9}}{s+3}\right]-e^{-5s}\left[\frac{\frac{19}{36}}{s}+\frac{\frac{1}{6}}{s^2}-\frac{\frac{7}{4}}{s+2}+\frac{\frac{11}{9}}{s+3}\right]$

 

Step 3: Take inverse Laplace transform we get

$$\begin{gathered} y\left(t\right)=2e^{-2t}-2e^{-3t}+\left[\frac{1}{36}u(t-1)+\frac{1}{6}(t-1)u(t-1)-\frac{1}{4}{e}^{-2(t-1)}u(t-1)+\frac{2}{9}{e}^{-3(t-1)}u(t-1)\right] \\ -\left[\frac{19}{36}u(t-5)+\frac{1}{6}(t-5)u(t-5)-\frac{7}{4}{e}^{-2(t-5)}u(t-5)+\frac{11}{9}{e}^{-3(t-5)}u(t-5)\right] \\ =2e^{-2t}-2e^{-3t}+\left[\frac{1}{36}+\frac{1}{6}(t-1)-\frac{1}{4}{e}^{-2(t-1)}+\frac{2}{9}{e}^{-3(t-1)}\right]u(t-1) \\ -\left[\frac{19}{36}+\frac{1}{6}(t-5)-\frac{7}{4}{e}^{-2(t-5)}+\frac{11}{9}{e}^{-3(t-5)}\right]u(t-5) \end{gathered}$$

Hence 

$\left.y\left(t\right)=2e^{-2t}-2e^{-3t}+\left[\frac{1}{36}+\frac{1}{6}(t-1)-\frac{1}{4}{e}^{-2(t-1)}+\frac{2}{9}{e}^{-3(t-1)}\right]u(t-1)-\left[\frac{19}{36}+\frac{1}{6}(t-5)-\frac{7}{4}{e}^{-2(t-5)}+\frac{11}{9}{e}^{-3(t-5)}\right]u(t-5)\right]$