Q14E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 1–19, use the method of Laplace transforms to solve the given initial value problem. Here x′, y′, etc., denotes differentiation with respect to t; so does the symbol D.

$\begin{array}{ccc}{\mathbf{\text{x}}}^{\mathbf{\text{''}}}\mathbf{\text{=y+u(t-1);}}& \mathbf{\text{x(0)=1,}}& {\mathbf{\text{x}}}^{\mathbf{\text{'}}}\mathbf{\text{(0)=0}}\\ {\mathbf{\text{y}}}^{\mathbf{\text{''}}}\mathbf{\text{=x+1-u(t-1);}}& \mathbf{\text{y(0)=0,}}& {\mathbf{\text{y}}}^{\mathbf{\text{'}}}\mathbf{\text{(0)=0}}\end{array}$

Step-by-Step Solution

Answer

The solution is 

$\begin{array}{l}x\left(t\right)=[1-cos(t-1\left)\right]u(t-1)+\frac{cost}{2}+\frac{sint}{2}+\frac{e^t}{2}-1\\ y\left(t\right)=\left[cos\right(t-1)-1]u(t-1)-\frac{cost}{2}-\frac{sint}{2}+\frac{e^t}{2}\end{array}$

Step 1: Given information

The differential equations are given as:

$\begin{array}{ccc}{x}^{\text{'}\text{'}}=y+u(t-1);& x\left(0\right)=1,& x^{\text{'}}\left(0\right)=0\\ y^{\text{'}\text{'}}=x+1-u(t-1);& y\left(0\right)=0,& y^{\text{'}}\left(0\right)=0\end{array}$

Step 2: Apply the Laplace transform

Given initial value equations are,

$\begin{array}{ccc}{x}^{\text{'}\text{'}}=y+u(t-1);& x\left(0\right)=1,& x^{\text{'}}\left(0\right)=\mathrm{0....}\left(\text{1}\right)\\ y^{\text{'}\text{'}}=x+1-u(t-1);& y\left(0\right)=0,& y^{\text{'}}\left(0\right)=\mathrm{0.....}\left(\text{2}\right)\end{array}$

Taking Laplace transform of equation first we get

$\begin{array}{c}{s}^{2}x\left(s\right)-s{x}^{\text{'}}\left(0\right)-x\left(0\right)=y\left(s\right)+\frac{e^{-s}}{s}\\ s^{2}x\left(s\right)-1=y\left(s\right)+\frac{e^{-s}}{s}\mathrm{....}\left(3\right)\end{array}$

Taking Laplace transform of equation second we get 

$\begin{array}{c}{s}^{2}y\left(s\right)-s{y}^{\text{'}}\left(0\right)-y\left(0\right)=x\left(s\right)+\frac{1}{s}-\frac{e^{-s}}{s}{s}^2\\ y\left(s\right)=x\left(s\right)+\frac{1}{s}-\frac{e^{-s}}{s}\\ x\left(s\right)=s^{2}y\left(s\right)-\frac{1}{s}+\frac{e^{-s}}{s}\mathrm{.....}\left(4\right)\end{array}$

Putting equation fourth into third we get 

$\begin{array}{c}{s}^{4}y\left(s\right)-s+s{e}^{-s}-1=y\left(s\right)+\frac{e^{-s}}{s}\\ (s^4-1)y\left(s\right)=\frac{e^{-s}}{s}+s-s{e}^{-s}+1\\ y\left(s\right)=\frac{e^{-s}}{s(s^4-1)}+\frac{s-s{e}^{-s}+1}{(s^4-1)}\\ =\frac{(1-s^2)e^{-s}}{s(s^4-1)}+\frac{s+1}{(s^4-1)}\end{array}$

Step 3: Take inverse Laplace

Taking partial fraction we get

$\begin{array}{c}y\left(s\right)=e^{-s}\left[\frac{s}{(s^2+1)}-\frac{1}{s}\right]+\frac{-s-1}{2(s^2+1)}+\frac{1}{2(s-1)}\\ =e^{-s}\left[\frac{s}{(s^2+1)}-\frac{1}{s}\right]+\frac{-s}{2(s^2+1)}-\frac{1}{2(s^2+1)}+\frac{1}{2(s-1)}\end{array}$

Taking inverse Laplace transform we get 

$y\left(t\right)=\left[cos\right(t-1)-1]u(t-1)-\frac{cost}{2}-\frac{sint}{2}+\frac{e^t}{2}$

Similarly solving equation fourth and third we get 

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

Taking partial fraction we get

$\begin{array}{c}x\left(s\right)=e^{-s}\left[\frac{1}{s}-\frac{s}{(s^2+1)}\right]+\frac{s+1}{2(s^2+1)}+\frac{1}{2(s-1)}-\frac{1}{s}\\ =e^{-s}\left[\frac{1}{s}-\frac{s}{(s^2+1)}\right]+\frac{s}{2(s^2+1)}+\frac{1}{2(s^2+1)}+\frac{1}{2(s-1)}\end{array}$

Taking inverse Laplace transform we get 

$x\left(t\right)=[1-cos(t-1\left)\right]u(t-1)+\frac{cost}{2}+\frac{sint}{2}+\frac{e^t}{2}-1$

Hence

$\begin{array}{l}x\left(t\right)=[1-cos(t-1\left)\right]u(t-1)+\frac{cost}{2}+\frac{sint}{2}+\frac{e^t}{2}-1\\ y\left(t\right)=\left[cos\right(t-1)-1]u(t-1)-\frac{cost}{2}-\frac{sint}{2}+\frac{e^t}{2}\end{array}$

Step 4: Conclusion

The final solution is 

$\begin{array}{l}x\left(t\right)=[1-cos(t-1\left)\right]u(t-1)+\frac{cost}{2}+\frac{sint}{2}+\frac{e^t}{2}-1\\ y\left(t\right)=\left[cos\right(t-1)-1]u(t-1)-\frac{cost}{2}-\frac{sint}{2}+\frac{e^t}{2}\end{array}$