Q27E - 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.

$$\begin{gathered} \mathit{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathit{z}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{z}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{3}\mathbf{t}}\mathit{u}\mathbf{(}\mathit{t}\mathbf{-}\mathbf{2}\mathbf{)} \\ \mathit{z}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{ }\mathit{z}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{3} \end{gathered}$$

 

Step-by-Step Solution

Answer

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

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

 

Step 1: Define Laplace Transform

The use of Laplace transformation is to convert differential equations into algebraic equations. The formula for Laplace transform is   

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

Where, F(s) = Laplace Transform

S = complex number

t = real number >=0 

t’ = first derivative of the function f(t)

 

Step 2: Apply Laplace transform

 

Given initial value problem

$z\text{'}\text{'}+3z\text{'}+2z=e^{-3t}u(t-2)$

where.$z\left(0\right)=2 and z\text{'}\left(0\right)=3$

Taking Laplace transform of initial value problem is

$\mathcal{L}z\text{'}\text{'}\left(s\right)+3\mathcal{L}z\text{'}\left(s\right)+2\mathcal{L}z\left(s\right)=\mathcal{L}\left[e^{-3t}u(t-2)\right]$

$$\begin{gathered} s^2\mathcal{L}z\left(s\right)-s\mathcal{L}z\left(0\right)-z\text{'}\left(0\right)+3s\mathcal{L}z\left(s\right)-3z\left(0\right)+2\mathcal{L}z\left(s\right)=\frac{e^{-2s}{e}^{-6}}{s+3} \\ {s}^2\mathcal{L}z\left(s\right)-2s+3+3s\mathcal{L}z\left(s\right)-6+2\mathcal{L}z\left(s\right)=\frac{e^{-2s}{e}^{-6}}{s+3} \\ \left(s^2+3s+2\right)\mathcal{L}z\left(s\right)-(2s+3)=\frac{e^{-2s}{e}^{-6}}{s+3} \end{gathered}$$

$$\begin{gathered} \mathcal{L}z\left(s\right)=\frac{2s+3}{\left(s^2+3s+2\right)}+\frac{e^{-2s}{e}^{-6}}{(s+3)\left(s^2+3s+2\right)} \\ =\frac{1}{s+1}+\frac{1}{s+2}+\frac{e^{-2s}{e}^{-6}}{(s+3)\left(s^2+3s+2\right)}\dots \left(1\right) \end{gathered}$$

 

Using partial fraction

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

Equation first becomes as,

$\mathcal{L}z\left(s\right)=\frac{1}{s+1}+\frac{1}{s+2}+\frac{\frac{e^{-2s}{e}^{-6}}{2}}{s+1}-\frac{e^{-2s}{e}^{-6}}{s+2}+\frac{e^{-2s{e}^{-6}}}{2}$

Taking inverse Laplace transform we get

$$\begin{gathered} z\left(t\right)=e^{-t}+e^{-2t}+\frac{1}{2}{e}^{-(t-2)}{e}^{-6}u(t-2)-e^{-2(t-2)}{e}^{-6}u(t-2)+\frac{1}{2}{e}^{-3(t-2)}{e}^{-6}u(t-2) \\ =e^{-t}+e^{-2t}+\frac{1}{2}{e}^{-t-4}u(t-2)-e^{-2t-2}u(t-2)+\frac{1}{2}{e}^{-3t}u(t-2) \end{gathered}$$

 

Hence,

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