Strategies for solving second order DEs with variable coefficients

Does anyone have any good videos/ article links for techniques to solve second order DEs with variable coefficients?

I mean, just as an example, a DE like:

$$x \frac{d^{2}y}{dx^{2}} - (2x-1) \frac{dy}{dx} + (x-1)y = 0$$

(this one is fairly simple to solve but the one I have in mind is tricker (for me anyway)).

If anyone's curious, the DE I have is:

$$\frac{d^2 y}{dx^2} + \frac{2(x^2 - 1)}{x^3} \frac{dy}{dx} + \frac{x^2 + 1}{x^6} y = 0$$

Are there any techniques to systematically find a general solution for DEs like this?

Thanks very much

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A good technique for these equations is the Frobenius method you assume the solution has the form of a power series and equate the coefficients of each power of $x$. mathworld.wolfram.com/FrobeniusMethod.html – Graham Hesketh Jul 20 '13 at 15:49

The best technique is a long, hard stare at the answer given by Wolfram Alpha: $$y(x)=\left(\frac{C_1}{x}+C_2\right)\exp\left\{-\frac{1}{2x^2}\right\}.$$

More seriously, when you have a linear 2nd order equation with rational coefficients, the first thing you need to look at is the structure of its singular points on the Riemann sphere. For example,

• if there are $3$ regular singularities, we are sure that the equation can be solved with Gauss hypergeometric functions,

• if there are $>3$ singular points, it is (almost) certain that the solutions cannot be written in a closed form.

• when we have $\leq 2$ singularities (like in your example), some of which are irregular (here $x=0$), then one should look in more details at the solution asymptotics to decide if there are any chances to solve the equation in terms of elementary or special functions.

In your case the only (irregular) singular point is $x=0$. To investigate the asymptotics, it is convenient to send it to infinity by the change of variables $t=1/x$, $u(t)=y(x)$ and then look for the asymptotic solution in the form $u(t)=t^{a}\exp P(t^b)$, where $P$ is a polynomial. But in your case this already gives the exact answer.

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