A pair of coupled second order linear ODEs. Where is my mistake?

Question

I'm given the following pair of coupled second order linear ODEs: $$x'' - k^2 (x+y) = 0 \\ y'' + k^2 (x+y) = 0$$ First, I did the following substitutions: $$x = Ae^{mt}, \ y = Be^{mt}$$ From this I got the following system of linear equations: $$(m^2 - k^2)A - k^2B = 0 \\ k^2A + (m^2 + k^2)B = 0$$ To get nontrivial solutions, the following determinant must be equal to zero: $$\begin{vmatrix} m^2-k^2 & -k^2 \\ k^2 & m^2+k^2 \end{vmatrix} = m^4 = 0$$ This means that we have $m=0$ four times. Thus, the solutions are $e^{0 \cdot t} = 1$, $te^{0 \cdot t} = t$, $t^2 e^{0 \cdot t} = t^2$, and $t^3 e^{0 \cdot t} = t^3$. From this the solution for $x$ is: $$x = A_1 + A_2t + A_3t^2 + A_4t^3$$ Since we started with a coupled system of ODEs, $A$ and $B$ aren't independent. By subbing $m=0$ into either one of the linear equations above, we get the following relationship between $A$ and $B$: $$(0-k^2)A - k^2B = 0 \\ B=-A$$ This leaves the solution for $y$ as: $$y = -A_1 - A_2t - A_3t^2 - A_4t^3$$ When I check my result by plugging $x$, $y$, $x''$, and $y''$ into the original pair of ODEs, the result is incorrect. Where did I go wrong with my solution, because it has worked when I've applied it to some other similar pairs of coupled ODEs.

Answer 1

Assuming you know the Laplace Transform. After transforming we get

$$ (s^2-k^2)X(s)-k^2 Y(s) = s x(0)+x'(0)\\ k^2 X(s) + (s^2+k^2)Y(s) = s y(0)+y'(0) $$

and solving for $X(s), Y(s)$ we get

$$ X(s) = \frac{x(0)}{s} + \frac{x'(0)}{s^2}+ k^2(x(0)+y(0))\frac{1}{s^3}+k^2(x'(0)+y'(0))\frac{1}{s^4}\\ Y(s) = \frac{y(0)}{s} + \frac{y'(0)}{s^2}- k^2(x(0)+y(0))\frac{1}{s^3}-k^2(x'(0)+y'(0))\frac{1}{s^4} $$

etc.

Answer 2

The solutions are of a much simpler form than you are trying to find.

$x''+y''=0$ and so $x+y=A+Bt$.

$x'' = k^2(A+Bt)$ therefore $x=C+Dt+k^2(\frac {A}{2}t^2+\frac {B}{6}t^3)$

Then $y=A-C+(B-D)t-k^2(\frac {A}{2}t^2+\frac {B}{6}t^3)$