I need to solve the following system of coupled 2nd order differential equations:
$$
\left\{
\begin{array}{c}
(m_1+m_2)Lx'' +m_2Ly''+(m_1+m_2)gx = 0 \\
Ly''+Lx''+gy = 0
\end{array}
\right.
$$
with $m_1$, $m_2$ and $L$ being constants.
I would really appreciate if someone could advise on the method that I could use. I tried making a substitution $x'=a$ and $y'=b$ but it did not work.
Thank you in advance!
Best Answer
First we reduce the number of parameters:
$$\left\{ \begin{array}{c} x'' +ay''+bx = 0 \\ x''+y''+by = 0 \end{array} \right. $$
$$a=\frac{m_2}{m_1+m_2}$$
$$b=\frac{g}{L}$$
Then we 'solve' the system as linear equations for $x''$ and $y''$:
$$x''=-\frac{b}{1-a} x+\frac{a b}{1-a} y$$
$$y''=\frac{b}{1-a} x-\frac{b}{1-a} y$$
Now introduce new functions for the first derivatives. Finally our system becomes:
$$ \begin{cases} x'=p \\ y'=q \\ p'=-\frac{b}{1-a} x+\frac{a b}{1-a} y \\ q'=\frac{b}{1-a} x-\frac{b}{1-a} y \end{cases}$$
This is a system of linear 1st order ODEs with constant coefficients and can be easily solved by the usual matrix methods.