solution to the nonlinear 2nd order d.e

[younginmoon]

Hello:

Can anyone help me solve with the following nonlinear 2nd order differential equation?

d^2 y/dx^2 (1+a(dy/dx)^2)=bx^c
(a,b & c are constants.)



Thank you.

younginmoon

[smallphi]

Luckily, this equation doesn't contain the function y(x) itself, only its derivatives. So denote y'(x) = f(x), you get an equation of first order for f(x), which is solvable by simplle integration. Then go back and solve y'(x) = f(x) for y(x).

[younginmoon]

Luckily, this equation doesn't contain the function y(x) itself, only its derivatives. So denote y'(x) = f(x), you get an equation of first order for f(x), which is solvable by simplle integration. Then go back and solve y'(x) = f(x) for y(x).

Thanks!
Starting with v=y', the original equation becomes
v + A v^3 = B x^C (A, B & C are constants) and the solution is composed of complimentary and particular integrals. But, how do you handle with the cubic term (v^3)
in both integrals? Or is there another solution method? (younginmoon)

[smallphi]

Cardano formulas for solving the qubic equation for v, but they are quite a mess.

[Mute]

Thanks!
Starting with v=y', the original equation becomes
v + A v^3 = B x^C (A, B & C are constants) and the solution is composed of complimentary and particular integrals. But, how do you handle with the cubic term (v^3)
in both integrals? Or is there another solution method? (younginmoon)


Don't forget the additive constant obtained when you integrate the first time:

v + \frac{a}{3}v^3 = \frac{b}{c+1}x^{c+1} + k

As for solving, you might try guessing that you can write the expression in the form (v + f(x))^2(v + g(x)) = 0, then expand and compare to the equation above to see if you can choose f(x) and g(x) such that this expression holds and expands to your original equation. (You might instead try the more general (v+f(x))(v+g(x))(v+h(x)) = 0, as there's really no obvious reason to expect a double root). There is of course no guarentee you can solve the problem this way, since solving for f, g and h might be just as hard if not harder than solving for v, but it might be worth a shot.

[Matthew Rodman]

You can solve the cubic v equation with Vieta's Substitution. (http://mathworld.wolfram.com/VietasSubstitution.html) {Wolfram.com link}