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I'm learning to get series solutions for differential equations.

My book says: If you have a second order differential equation of the type:

$a_1(x) y'' + a_2(x)y' + a_3(x)y = 0$

we should first divide it by $a_1$, and then check for singular points, when all the terms are polynomials, the diff. equation will probably not have a solution on the point where $a_1$ is 0.

My question is, why divide everything by $a_1$? Can't we just check if $a_1, a_2$ and $a_3$ are analytical in $x_0$ and just get two linear independent solutions with Taylor Series?

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The problem is that where $a_1(x)=0$, your equation is not really second order. It doesn't mean that it doesn't have solutions, but fact is that the zeroes of $a_1$ are singular points, i.e., points where usually bad things happen to the solution.

In most cases, $a_1$ has just a few zeroes, and one simply works within the intervals determined by those zeros. Within those intervals, dividing by $a_1$ is fine.

As you say, if your goal is to solve by using power series, dividing by $a_1$ may complicate things. Note, though, that if you will have $a_1$ as a nontrivial power series, you will need to multiply that series with that of $y''$, and in most cases you won't get anything remotely pretty. The most common cases occur when $a_1$ is a power of $x$ or another equally simple expression.

If you are using a method other than power series, like variation of parameters, most of the time the method is phrased with the equation in a form where the first term is $y''$ alone.

Finally, to answer your title question: in general you will not lose solutions because the singular points are isolated and you look for solutions on intervals. If $a_1$ were to be zero on an interval (in which case it cannot be analytic), then you would indeed lose solutions by dividing. All methods I know wouldn't apply in that situation, though.

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  • $\begingroup$ Glad I could help! $\endgroup$ – Martin Argerami Apr 12 '18 at 4:03

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