Second-order Coupled O.D.E.s with constant coefficients

[ab959]

Homework Statement

I am trying to solve a system of two coupled ODEs. I am interested in an analytic solution if that is possible. I know it will be messy.

\frac{\partial^2 U_1}{\partial x^2}+a_1\frac{\partial U_1}{\partial x^2}+b_1 U_1 = c_1 U_2

\frac{\partial^2 U_2}{\partial x^2}+a_2\frac{\partial U_2}{\partial x^2}+b_2 U_2 = c_2 U_1

The Attempt at a Solution

I have attempted a few methods although I haven't pursued them too hard as I don't want to head down the wrong path. I found that a Laplace transform was too hard to invert. I tried to represent it as a system of 4 first order equations however the inversion was very very messy. Finally, the method I was thinking of employing was to treat the U_2 in the first equation as an arbitrary function and then exploiting the symmetry to find a solution of U_1/U_2 in terms of something in the form of:

U_{1,2}=... + \int^x_{x_0} U_{2,1}(\eta)e^{(x-\eta)k}d\eta

, where k is a variable in terms of a_1 and b_2. I do not know if this will work and have not done much in this area.

Does anyone have any suggestions as I am beginning to feel a bit out of my depth. Thanks.

 

[quantumdude]

Solve the first equation for U_2 and plug it into the second equation.

\frac{\partial^2 U_1}{\partial x^2}+a_1\frac{\partial U_1}{\partial x^2}+b_1 U_1 = c_1 U_2

\frac{\partial^2 U_2}{\partial x^2}+a_2\frac{\partial U_2}{\partial x^2}+b_2 U_2 = c_2 U_1

Why is there a "\partial x^2" in the second term of the left hand side of each equation?