Solve system differential equation using Laplace transform

Question

Question : Solve system differential equation using Laplace transform :

$\cases{x'-3x+2y= \sin t\\4x-y'-y=\cos t\\x(0)=0,y(0)=0}$

My try :

Take Laplace we find :

$\cases{(s-3)X(s)=\frac{1}{s^2+1}-2Y(s)\\(s+1)Y(s)=4X(s)-\frac{s}{1+s^2}}$

What's the fast way to solving this system :

My work :

$Y(s)=4\frac{X(s)}{1+s}-\frac{s}{(1+s)(1+s^2)}$

we find : $X(s)=\frac{1-s}{(1+s^2)(s^2-2s+5)}$

Please If any one have ideas to solving above system tell because this killed my time in exams

Answer 1

First, we note that $X(s)=\frac{1-s}{(1+s^2)(s^2-2s+5)}$ has partial fraction decomposition: $$ \frac{s-5}{10\left( {{s}^{2}}-2s+5\right) }-\frac{s-3}{10\left( {{s}^{2}}+1\right) }. $$ Next, simply look up in the Inverse Laplace transform table.

Answer 2

Well, we have the following coupled DE's:

$$ \begin{cases} \text{x}'\left(t\right)-3\cdot\text{x}\left(t\right)+2\cdot\text{y}\left(t\right)=\sin\left(t\right)\\ \\ 4\cdot\text{x}\left(t\right)-\text{y}'\left(t\right)-\text{y}\left(t\right)=\cos\left(t\right)\\ \\ \text{x}\left(0\right)=0\\ \\ \text{y}\left(0\right)=0 \end{cases}\tag1 $$

Using Laplace transform, and the table of selected Laplace transforms we can transform equation $(1)$ to the s-domain:

$$ \begin{cases} \text{s}\cdot\text{X}\left(\text{s}\right)-\text{x}\left(0\right)-3\cdot\text{X}\left(\text{s}\right)+2\cdot\text{Y}\left(\text{s}\right)=\frac{1}{1+\text{s}^2}\\ \\ 4\cdot\text{X}\left(\text{s}\right)-\text{s}\cdot\text{Y}\left(\text{s}\right)-\text{y}\left(0\right)-\text{Y}\left(\text{s}\right)=\frac{\text{s}}{1+\text{s}^2}\\ \\ \text{x}\left(0\right)=0\\ \\ \text{y}\left(0\right)=0 \end{cases}\tag2 $$

Using the initial conditions, we get:

$$ \begin{cases} \text{s}\cdot\text{X}\left(\text{s}\right)-3\cdot\text{X}\left(\text{s}\right)+2\cdot\text{Y}\left(\text{s}\right)=\frac{1}{1+\text{s}^2}\\ \\ 4\cdot\text{X}\left(\text{s}\right)-\text{s}\cdot\text{Y}\left(\text{s}\right)-\text{Y}\left(\text{s}\right)=\frac{\text{s}}{1+\text{s}^2}\\ \end{cases}\tag3 $$

Solving for $\text{X}\left(\text{s}\right)$ and $\text{Y}\left(\text{s}\right)$, we get:

$$ \begin{cases} \text{X}\left(\text{s}\right)=\frac{\frac{1}{1+\text{s}^2}-2\text{Y}\left(\text{s}\right)}{\text{s}-3}\\ \\ \text{Y}\left(\text{s}\right)=-\frac{\frac{\text{s}}{1+\text{s}^2}-4\text{X}\left(\text{s}\right)}{\text{s}+1}\\ \end{cases}\tag4 $$

Subsituting them into each other gives:

1. $$\text{X}\left(\text{s}\right)=\frac{\frac{1}{1+\text{s}^2}-2\left(-\frac{\frac{\text{s}}{1+\text{s}^2}-4\text{X}\left(\text{s}\right)}{\text{s}+1}\right)}{\text{s}-3}\tag5$$
2. $$\text{Y}\left(\text{s}\right)=-\frac{\frac{\text{s}}{1+\text{s}^2}-4\left(\frac{\frac{1}{1+\text{s}^2}-2\text{Y}\left(\text{s}\right)}{\text{s}-3}\right)}{\text{s}+1}\tag6$$

Steps you need to take: solve both the above equations for the function that your looking for and perform the inverse Laplace transform on them.