# Thread: Second order D.E involving substitution

1. ## Second order D.E involving substitution

I'm trying to solve the second order D.E: sin(theta)y'' -cos(theta)y' + 2ysin(theta)^3 = 0 using the substitution x = cos(theta). According to Wolfram Alpha, there exists a trigonometric solution but my working does not lead me towards it. Here is what I have:

Please excuse the bad handwriting - it seems to be exacerbated by scanning my working in!

2. Your working seems fine all the way down to

$\dfrac{d^{2}y}{dx^{2}}=-2y.$

Your integration of this DE might be correct, but that's the long way to go about it. You have to invert a rather messy function there, which might not be the easiest thing to do.

This is just a second-order ODE with constant coefficients, right? Why not just do guess-and-check? Surely, if you're doing a substitution this complicated, you know what the solution to the above DE is.

3. Originally Posted by Ackbeet
Your working seems fine all the way down to

$\dfrac{d^{2}y}{dx^{2}}=-2y.$

Your integration of this DE might be correct, but that's the long way to go about it. You have to invert a rather messy function there, which might not be the easiest thing to do.

This is just a second-order ODE with constant coefficients, right? Why not just do guess-and-check? Surely, if you're doing a substitution this complicated, you know what the solution to the above DE is.
Ah yes, using the usual method of trying e^kx as a trial solution works fine. I get y=k0*cos(root(2)cos(theta)) + k1*sin(root(2)cos(theta)). This agrees with wolfram alpha, though the computer generated solution is convoluted. I think I was over-tempted by the fact I could solve by integration!

4. That looks right to me.