Q28E - 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.

$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{=}\mathit{t}\mathit{u}\mathbf{(}\mathit{t}\mathbf{-}\mathbf{2}\mathbf{)} \\ \mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{ }\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1} \end{gathered}$$

 

Step-by-Step Solution

Answer

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

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

 

Step 1: Define Laplace Transform

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

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

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

$y\text{'}\text{'}+5y\text{'}+6y=tu(t-2)$

where. $y\left(0\right)=0 and y\text{'}\left(0\right)=1$

Taking Laplace transform of initial value problem is

$\mathcal{L}y\text{'}\text{'}\left(s\right)+5\mathcal{L}y\text{'}\left(s\right)+6\mathcal{L}y\left(s\right)=\mathcal{L}\left[tu\right(t-2\left)\right]$

$$\begin{gathered} s^2\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+5s\mathcal{L}y\left(s\right)-5y\left(0\right)+6\mathcal{L}y\left(s\right)=e^{-2s}\left[\frac{1}{s^2}+\frac{2}{s}\right] \\ {s}^2\mathcal{L}y\left(s\right)-0-1+5s\mathcal{L}y\left(s\right)-0+6\mathcal{L}y\left(s\right)=e^{-2s}\left[\frac{1+2s}{s^2}\right] \\ \left(s^2+5s+6\right)\mathcal{L}y\left(s\right)-1=e^{-2s}\left[\frac{1+2s}{s^2}\right] \end{gathered}$$

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

 

Using partial fraction

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

Equation first becomes as

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

 

Taking inverse Laplace transform we get

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

 

Hence,

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