# General solution of a linear system (differential equations)

1. Oct 29, 2009

### wtrow

1. The problem statement, all variables and given/known data
x''+13y'-4x=6sint , y''-2x'-9y=0

3. The attempt at a solution
I am not really sure how to solve this completely, but I have done this so far:

(D^2-4)x + 13Dy - 6sint = 0 , (D^2-9)y - 2Dx = 0

then I hit a brick wall. Any help would be appreciated, thanks.

2. Oct 30, 2009

### Staff: Mentor

Do you know about eigenvectors, eigenvalues, and matrix diagonalization? The system you have here is an example of a coupled linear system. With suitable substitutions it can be converted from a system of two second-order (nonhomogeneous) differential equations into a system of four first-order differential equations, also nonhomogeneous.

Let's ignore the 6sint term for a while, which makes the system homogeneous. This substitution can be used to get to the four first-order equations:
y1 = y
y2 = y'
y3 = x
y4 = x'

With these substitutions, your system of two equations can be rewritten as:
y1' = y2
y2' = 4y1 - 13 y4
y3' = y4
y4' = 2y2 + 9y3

This system of equations can be written in matrix form as y' = Ay,
with
$$A~=~\left[ \begin{array}{c c c c} 0&1&0&0\\ 4&0&0&-13\\ 0&0&0&1\\ 0&2&9&0\\ \end{array} \right]$$

The solution of the matrix differential equation y' = Ay is y = etAc, where c is a vector of constants. Where the eigenvalues, eigenvectors, and matrix diagonalization come in, is that it is much easier to evaluate e raised to a matrix power if the matrix is diagonal.

I hope some of these ideas are familiar to you. Yours is not a simple problem, and there is still the question of dealing with the nonhomogeneous system, which is not that more complicated if you understand what I've laid out here.

3. Oct 30, 2009

### HallsofIvy

Mark44 is completely correct but, but from what you have written, I suspect you may not be ready for that method.

Treat these equations as algebraic equation for x and y and eliminate one of them. For example, if you "multiply" the first equation by 2D (actually you are differentiating the equation and multiplying by 2) you get 2D^2(D^2- 4)x+ 26D^2y= 6D(sin t)= 6cos t.
If you "multiply" the second equation by D^2- 4 (actually Differentiate the entire equation twice and subtract the equation from that) you get (D^2-4)(D^2-9)y- 2D(D^2-4)x= 0.

Adding the two equations eliminates x and gives you a fourth order equation for y.

4. Oct 30, 2009

### Staff: Mentor

HallsOfIvy, Thanks for jumping in on this with a simpler (and therefore better) approach.

To clarify, it looks like you are "multiplying" the first equation by the 2D2 operator.