Simplifying system of two-variable differential equations

Question

How can I make progress with a system of differential equations of the following type: (the variables are $y_1$ and $y_2$ and all constants definitve positive). $$ \begin{align} \ddot{y}_1+by_1 &= -a\dot{y}_2-k \tag{1}\\ \ddot{y}_2+by_2 &= a\dot{y}_1+k \tag{2} \end{align} $$ Is there an easy way to detach them from one another? so we'd be able to deal with one variable at a time. (in a higher order differential equation possibly?)

Answer 1

I don't know if you are familiar with the Laplace Transformation, but whenever I have trouble dealing with differential equations, I use it to simplify it. Since you have no start values, you can just assume all of them are 0 so you get:

$s²Y_1+bY_1 = -asY_2 - k$

$s²Y_2+bY_2 = asY_1+k$

Where $Y_i(s)$ is the lamplace transformed of $y_i(t)$. Note that taking a differential in the time domain will result in multiplication with $s$ in the $s$ domain.

Now it should be pretty simple to solve for $Y_1$ and $Y_2$ and you get:

$Y_2(s⁴+2s²b+b²)=Y_2(s²+b)²=aY_1(S²+b)+k(s²+b)=a(-aY_2-k)+k(s²+b)=-a²Y_2-ak+ks²+kb$ If you transform this back you will get: $y_2^{(4)}+2y_2^{(2)}b+y_2(a²+b²)=k(b-a)$. You will also get something similar for $y_1$. I know this isn't the most elegant method, but it works all the time and if you need some more formal proof how you got it, you can at least know if you are heading in the right direction. Hope I didn't make some mistake there and that it will help you a bit.

Answer 2

This is a brief answer. Introduce $z_1 = \dot{y}_1$ and $z_2=\dot{y}_2$, so that you would have first order equations for 4 variables. These equations are linear, so I would recommend to use Laplace transform to solve them directly (this is what MAXIMA does, for example). Alternatively, you would need to diagonalise the matrix: $A=VDV^{-1}$ where $D$ is the diagonal matrix with eigenvalues on diagonals and $V$ is the matrix of eigenvectors (this assumes there is no problem with geometric multiplicity of eigenvalues, otherwise $D$ won't be diagonal). Then, $z=V^{-1}x$ and solve $\dot{z}=Dz$, which is elementary. Then, $x=Vz$. The difficulty here is to find the decomposition $A=VDV^{-1}$ (and hence I originally thought SVD can do side-step this problem, but then it doesn't work out there as nicely).