Hi,
> y' + P(x)y^2 + Q(x) = 0. (1)
> Is this equation solvable in general?
- At lest it's linearized. Do the change
y = u'/[P(x)u].
Performing this change into (1) we get
P(x)u'' - P'(x)u' + P^2(x)Q(x)u = 0,
or
u'' - (P'/P)u' +PQu = 0,
or
u'' - (logP)'u' PQu = 0,
linear but lifting to 2nd order and quite general, but afortunately homogeneus(we must to pay some sacrifice). Singularities at the roots of P(x) (at least P'/P be defined at those roots which is seldom).
Now it's your turn by investigating that horripilant ODE in a good math handbook on differential eqs., and you also need boundary (or initial) conditions (look at the change above). For sure the general solution are special functions (orthogonal and complete on the null space function) as e.g. Bessel, Hankel, Hermite error, hypergeometric functions, etc.(there are a lot!), or also special polynomial series as e.g. Bernoulli, Chebyshev, La Guerre, Legendre, Neumann, etc. (there are a lot too!), but also Frenel, Elliptic, Hyper-Elliptic integrals, etc., all depending on the form of P(x) & Q(x) and the above mentioned conditions.
Have good luck! (and don't be afraid please!)
perucho |
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