Solving a Two-Level System of Differential Equations with Laplace Transform

[indigojoker]

Homework Statement

How would I solve a two-level system such as this:

$$y_1'=-ay_1+by_2$$

$$y_2'=-ay_2-by_1+c$$

where a b and c are constants.

Homework Equations

Laplace transform

The Attempt at a Solution

I guess my question is what method would I use to solve this? As in should I solve for y1 in terms of y2 and then plug that back into the second equation? Not too sure about the approach.

 

[Glass]

I would do exactly as you suggested - solve for y2 in terms of y1 and sub back in. You will get a 2nd order ODE, but it is not too difficult. Moreover, if your unsure about the answer, you can always plug your solution back into the equation to ensure your solution satisfies the DE (although this doesn't show you have found all possible solutions). That's one of the great things about them - it's pretty easy to check if your solution is correct.

 

[HallsofIvy]

Homework Statement

How would I solve a two-level system such as this:

$$y_1'=-ay_1+by_2$$

$$y_2'=-ay_2-by_1+c$$

where a b and c are constants.

Homework Equations

Laplace transform

The Attempt at a Solution

I guess my question is what method would I use to solve this? As in should I solve for y1 in terms of y2 and then plug that back into the second equation? Not too sure about the approach.

I don't see any way to "solve for y1 in terms of y2" from the first equation because of the $y_1'$ term, nor do I see any reason to use the Laplace transform for such a simple problem. There are, however, several methods you could use. One is to convert from two first order equations to one second order equation: Differentiate the first equation to get $y_1"= -ay_1'+ by_2'$. Now replace that $y_2'$ with it's expression in the second equation: $y_1"= -ay_1'+ b(-cy_1+ dy_2)= -ay_1'- bcy_1+ d(by_2)$. From the first equation again, $by_2= y_1'+ ay_1$ so the equation becomes $y_1"= -ay_1'-bcy_1+ dy_1'+ ady_1$ or y_1"+(a-d)y_1'+ (ad- bc)y_1= 0[/itex]. Solve that equation for $y_1$ and then use $by_2= y_1'+ ay_1$ to solve for $y_2$.

Another way is to write it as a matrix equation:
If you let
$$Y= \left[\begin{array}{c}y_1 \ y2 \end{array}\right]$$
then your equation becomes
$$Y'= \left[\begin{array}{cc}a & b \ c & d\end{array}\right]Y$$
You can solve that by finding the eigenvalues and corresponding eigenvectors of the matrix.