# A set of two 2nd order linear ode

1. Jan 20, 2010

### omyojj

I'm doing an undergrad research job..
I have encountered the following coupled 2nd order linear ODE with constant coefficients a, b, c, d...
\begin{align} \frac{d^2 y_1}{d x^2} + a^2y_1 & = -c y_2 \\ \frac{d^2 y_2}{d x^2} - b^2y_2 & = -d y_1 \end{align}

In addition, I would like to impose constraints
$$\frac{d y_1}{d x}|_{x=0} = \frac{d y_2}{d x}_{x=0} = 0$$
(y_1, y_2 are even in x)

if c were zero..the solution would be simple..
\begin{align} y_1 & = A \cos(ax) \\ y_2 & = B \cosh(bx) + C \cos(ax) \end{align}
but i cannot proceed further when c is not zero..
did I fall into the silly trap??
plz help me find the solutions..

2. Jan 20, 2010

### elibj123

Re-write this vector form:

$$\vec{y}''=A\vec{y}$$

with y=[y1 y2]
and A=[-a^2 -c; -d b^2 ]

Like in an ordinary constant coefficient ODE, we suggest solutions of the form
$$\vec{y}=\vec{v}e^{sx}$$

Substituting into the equation gives:

$$s^{2} \vec{v}e^{sx}=A\vec{v}e^{sx}$$

or

$$(A-s^{2}I)\vec{v}=0$$

For this algebric equation to have a non-trivial solution, the determinant of the matrix must equal to zero. You might recognize this as searching for eigenvalues, only here the eigenvalues are $$\lambda=s^{2}$$.

In the most simple case, you'll get two real positive eigenvalues with one eigenvector which,
for which you will have four basic solutions:

$$\vec{y}=K_{11}\vec{v_{1}}e^{\sqrt{\lambda_{1}}x}+K_{12}\vec{v_{1}}e^{-\sqrt{\lambda_{1}}x}+K_{21}\vec{v_{2}}e^{\sqrt{\lambda_{2}}x}+K_{22}\vec{v_{2}}e^{-\sqrt{\lambda_{2}}x}$$

Off course other cases, will need more algebric manipulation, to pull out a real solution (f.e. if you get complex eigenvalues, your solutions will be periodic)