Kelton Bailey

2022-09-06

Help with coupled differential equations

I have these two coupled differential equations

$$a\ddot{x}+b\stackrel{\u20db}{x}+c\mathrm{tan}(x+y)\dot{x}=0,$$

$$a\ddot{y}-b\stackrel{\u20db}{y}-c\mathrm{tan}(x+y)\dot{y}=0$$

with $a,b,c=const.$

Now I basically have no experience at all with more advanced differential equations. Are these even solvable? I guess answering this type of question is already a science to itself, but maybe you can share some intuition under which circumstances one can hope to find a solution? What are possible methods or tricks?

I have these two coupled differential equations

$$a\ddot{x}+b\stackrel{\u20db}{x}+c\mathrm{tan}(x+y)\dot{x}=0,$$

$$a\ddot{y}-b\stackrel{\u20db}{y}-c\mathrm{tan}(x+y)\dot{y}=0$$

with $a,b,c=const.$

Now I basically have no experience at all with more advanced differential equations. Are these even solvable? I guess answering this type of question is already a science to itself, but maybe you can share some intuition under which circumstances one can hope to find a solution? What are possible methods or tricks?

Iademarco11

Beginner2022-09-07Added 10 answers

Step 1

The only exact solution I can find is

$$x={c}_{1},\phantom{\rule{thickmathspace}{0ex}}y={c}_{2},\phantom{\rule{thickmathspace}{0ex}}\text{such that}{c}_{1}+{c}_{2}\ne \frac{(2n+1)\pi}{2}\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}n\in \mathbb{Z}.$$

You need the condition on the sum to prevent evaluating the tangent function at one of its vertical asymptotes.

Step 2

It's possible to play games by solving each equation for the tangent function and then equating the two. What's nice about that: you get all the x's on one side and all the y's on the other and you can immediately reduce the order of the equation by one (corresponding to the constant solution above). What's not nice about that: you're down to only one equation, an equation you cannot directly integrate even if you try to do something fancy like make one side look like the derivative of a quotient. While the coupling of these ODE's looks a bit weak, it's still strong enough to resist analytical techniques.

The only exact solution I can find is

$$x={c}_{1},\phantom{\rule{thickmathspace}{0ex}}y={c}_{2},\phantom{\rule{thickmathspace}{0ex}}\text{such that}{c}_{1}+{c}_{2}\ne \frac{(2n+1)\pi}{2}\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}n\in \mathbb{Z}.$$

You need the condition on the sum to prevent evaluating the tangent function at one of its vertical asymptotes.

Step 2

It's possible to play games by solving each equation for the tangent function and then equating the two. What's nice about that: you get all the x's on one side and all the y's on the other and you can immediately reduce the order of the equation by one (corresponding to the constant solution above). What's not nice about that: you're down to only one equation, an equation you cannot directly integrate even if you try to do something fancy like make one side look like the derivative of a quotient. While the coupling of these ODE's looks a bit weak, it's still strong enough to resist analytical techniques.

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