How Do You Solve a System of Second Order ODEs with Matrix Methods?

[c0der]

Homework Statement

Solve:

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

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

Homework Equations

Everything required is in (1) above

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?

 

[Mugged]

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

 

[pasmith]

Homework Statement

Solve:

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

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

Homework Equations

Everything required is in (1) above

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

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

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?

 

[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