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?)

  • $\begingroup$ In addition to the answer below, another approach is to solve the first equation for $y'_2$, take the first, second and third derivative. Then take the derivative of the second equation and substitute and you will have a fourth order equation for $y_1$. Solve that and use to find $y_2$. Also note, you may need to do case work for $a = 0$. $\endgroup$
    – Moo
    Jan 23, 2018 at 13:49
  • $\begingroup$ @Moo Great! would you be so kind to post it as an answer? It will also be useful for future readers. $\endgroup$
    – user929304
    Jan 23, 2018 at 13:56

2 Answers 2


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.

  • $\begingroup$ Generally I won't say that $y_i(0)=0$ if it is not given but introduce new constant $c_i$ and set $y_i(0)=c_i$ to solve for general form $\endgroup$
    – ℋolo
    Jan 23, 2018 at 14:23
  • 1
    $\begingroup$ @Holo you are right, you can't just assume it like that, but it simplifies the calculation and what's important, you get a picture of what you need to get at the end, so you should know which steps to take, to get to a similar result. $\endgroup$
    – CryoDrakon
    Jan 23, 2018 at 15:00

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).

  • $\begingroup$ Thank you very much, indeed these are along the hints I was hoping for! For the alternative method, I lost you a bit :( what is SVD here? and how does it differ from the first solution you suggested? $\endgroup$
    – user929304
    Jan 23, 2018 at 13:34
  • $\begingroup$ @user929304 It's en.wikipedia.org/wiki/Singular-value_decomposition for the matrix $A$ in equations $\dot{x}=Ax$. The idea is to (linearly) change the variables in such a way that the matrix becomes diagonal and then express the solution via old variables. SVD is just a technique of doing it efficiently. $\endgroup$
    – mobiuseng
    Jan 23, 2018 at 13:45
  • $\begingroup$ @user929304 Scrap SVD - I was thinking about something else. You still can do matrix decomposition, but it will be more complicated. I will edit the answer. $\endgroup$
    – mobiuseng
    Jan 23, 2018 at 13:58

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