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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!

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1 Answer 1

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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.

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