# System of second order ODEs

1. Mar 16, 2014

### c0der

1. The problem statement, all variables and given/known data

Solve:

[ d^2y1/dx^2 ] = [ a -a ] [ y1 ]
[ d^2y2/dx^2 ] [ -a a ] [ y2 ]

A = [ a -a ]
[ -a a ]

2. Relevant equations

Everything required is in (1) above

3. The attempt at a solution

Reduce to 1st order system

M = [ 0 I ]
[ A 0 ]

Hence, M =
[ 0 0 1 0 ]
[ 0 0 0 1 ]
[ a -a 0 0 ]
[ -a a 0 0 ]

The eigenvalues of M are 0, 0, √2a and -√2a
The eigenvectors are [ 1 1 ], [ 1 1 ], [ 1 -1] and [ 1 -1]

Hence the general solution is (for y only):

y(x) = A*e^(0x)*[ 1 1 ]T + B*e^(0x)*[ 1 1 ]T + C*e^(√2a*x) [ 1 -1 ]T +
D*e^(-√2a*x) [ 1 -1 ]T

So:

y1 = A + B + C*cosh(√(2a)*x) + D*sinh(√(2a)*x)
y2 = A + B - C*cosh(√(2a)*x) - D*sinh(√(2a)*x)

Is this correct?

2. Mar 16, 2014

### Mugged

The way to check if thats correct is to plug your answer back into the ODE system.

3. Mar 16, 2014

### pasmith

Do you not recall that the solution of $u'' = 0$ is $u(x) = Ax + B$?

4. Mar 16, 2014

### c0der

Yes, ok. I see you mean because I have repeated roots, it's Ae^0(x) + Bxe^(0x). Even the above without A and B constants is a solution to the system. However, since I have 4 boundary conditions, I thought I'd keep them in there