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I'm looking for the solution method for solving an equation like the following:

$$a(x)y''+b(x)y'+c(x)y=g(x)$$

The solution methods I have come across are somewhat unsatisfactory because they tend to be for special cases of the above equation (for example, if the $y$ term is missing, we can convert it to a first order equation and use the integrating factor method).

I'd like to know a $\it general$ way to attack such a problem, if there is one.

Joebevo
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2 Answers2

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The general solution is variation of constants, which solves the problem $$ a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \ldots + a_1(x)y' = b(x). $$

Here we may as well assume $a_n(x) = 1$: if $a_n(x)=0$ then we reduce to the lower-order case, and otherwise we can divide the equation by $a_n(x)$. The general solution is given by the variation of constants formula: see http://en.wikipedia.org/wiki/Variation_of_parameters#Description_of_method.

This does not necessarily yield an explicitly computable answer or a solution in terms of elementary functions, but in principle the method always yields an exact answer; in practice the method reduces everything to quadrature, so it's good enough for numerics.

See the first section here for the Laplace transform applied to linear $n$-th order ODEs: http://en.wikipedia.org/wiki/Laplace_transform_applied_to_differential_equations.

Gyu Eun Lee
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  • Variation of constants only gives you the solution of the inhomogeneous equation one you know how to solve the homogeneous one. However, it tells nothing about the latter problem. – Start wearing purple Apr 06 '14 at 09:55
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    True. However, the homogeneous equation is solvable (perhaps not explicitly) under relatively mild conditions. This fact sometimes allows you to use the variation of constants formula in a more abstract proof, for instance existence of periodic solutions to quasilinear ODEs. I am not claiming that either problem is solvable in elementary terms, but in principle variation of constants gives the unique solution to the general equation given some weak constraints. – Gyu Eun Lee Apr 06 '14 at 10:02
  • Sorry, this is completely wrong: "solvable under relatively mild conditions". Otherwise please give me, e.g. the general solution of the (homogeneous) Heun's differential equation. I will award for such solution (quadratures would suffice) a 10K bounty. – Start wearing purple Apr 06 '14 at 10:10
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    You're mistaking my words. By "solvable" I mean "a solution exists." – Gyu Eun Lee Apr 06 '14 at 10:17
  • Ah ok, but the question is not about existence of solution. As far as I understand, it is about its explicit construction. – Start wearing purple Apr 06 '14 at 10:19
  • That's not how I read it. And variation of constants is, provided you can find a basis for the solution space of the homogeneous equation, an algorithm for the explicit construction of a solution. Besides which, as I have mentioned, the existence of solutions to the homogeneous equation under mild conditions makes the variation of constants formula useful in the abstract setting as well, so it would be unwise to discount the method completely. (Perhaps we're disagreeing on just how "explicit" a solution needs to be?) – Gyu Eun Lee Apr 06 '14 at 10:22
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In general, this equation can not be solved in terms of elementary functions nor classical special functions. By this I mean the homogeneous counterpart, since once the homogeneous equation is solved, the nonhomogeneous one can also be solved, at least in quadratures.

Solvable cases correspond to particular choices of $a(x),b(x),c(x)$. One way to understand if you are in such a special situation is to analyze the structure of singular points of the equation in the extended complex plane of $x$. This is an algorithmic procedure.

For example, if you have only 3 regular singular points (like here), the equation can be reduced to the hypergeometric one by a simple change of variables. When their number is greater than $3$, the equation is non-solvable, at least for generic coefficients. In the presence of irregular singularities, there is a similar procedure.