Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I solve this ODE:


Can you please also show the derivation.

share|cite|improve this question
What did you try? – timur Aug 31 '12 at 19:36
I think too many cases should be divided. Does there any constants have some restrictions for not equal to some values? – doraemonpaul Oct 29 '12 at 0:18

I don't think you'll find an "elementary" solution in general. Maple finds a rather complicated solution involving hypergeometric functions: $$\displaystyle S\, := \,y \left( x \right) ={\it \_C1}\,{\mbox{$_2$F$_1$}(1/2\,{\frac {-d+B+ \sqrt{{d}^{2}+ \left( -2\,B-4 \right) d+{B}^{2}}}{d}},1/2\,{\frac {-d+B- \sqrt{{d}^{2}+ \left( -2\,B-4 \right) d+{B}^{2}}}{d}};\,-1/2\\ \mbox{}\, \left( A-B \sqrt{-{\frac {C}{d}}} \right) {d}^{-1} \left( \sqrt{-{\frac {C}{d}}} \right) ^{-1};\,1/2\, \left( C- \sqrt{-{\frac {C}{d}}}xd \right) {C}^{-1})}+{\it \_C2}\, \left( C- \sqrt{-{\frac {C}{d}}}xd \right) ^{1/2\, \left( \left( -B+2\,d \right) \sqrt{-{\frac {C}{d}}}+A \right) {d}^{-1} \left( \sqrt{-{\frac {C}{d}}} \right) ^{-1}}{\mbox{$_2$F$_1$}(1/2\, \left( d \sqrt{-{\frac {C}{d}}}- \sqrt{{d}^{2}+ \left( -2\,B-4 \right) d+{B}^{2}} \sqrt{-{\frac {C}{d}}}+A \right) {d}^{-1} \left( \sqrt{-{\frac {C}{d}}}\\ \mbox{} \right) ^{-1},1/2\, \left( d \sqrt{-{\frac {C}{d}}}+ \sqrt{{d}^{2}+ \left( -2\,B-4 \right) d+{B}^{2}} \sqrt{-{\frac {C}{d}}}+A \right) {d}^{-1\\ \mbox{}} \left( \sqrt{-{\frac {C}{d}}} \right) ^{-1};\,1/2\, \left( \left( 4\,d-B \right) \sqrt{-{\frac {C}{d}}}+A \right) {d}^{-1} \left( \sqrt{-{\frac {C}{d}}} \right) ^{-1};\,1/2\, \left( C- \sqrt{-{\frac {C}{d}}}xd \right) {C}^{-1})} $$

(I used $d$ instead of $D$ because $D$ has a special meaning in Maple)

share|cite|improve this answer

Frobenius method is the most general method I know for this case. Assume your solution is of the form $y = x^r\sum_{n=0}^\infty a_n x^n$, plug it in, and solve for $r$ and $a_n$. You should get two $r$, say $r_1$ and $r_2$. If $r_1 - r_2$ is not an integer, you already have two linearly independent solutions. If $r_1 - r_2$ is an integer, then there are several ways to get two solutions. The simplest theoretical method would be reduction of order. There are, however, special ways tailored for Frobenius method. This seems like a good reference.

share|cite|improve this answer
Actually $r=0$ because (if $C \ne 0$) $0$ is a regular point. You'll get a not-particularly-nice three-term recurrence for $a_n$. However, you can get a two-term recurrence for the coefficients if you expand around one of the singular points, which are the roots of $C+Dx^2$. That's basically what Maple does to get its solution in terms of hypergeometric functions. – Robert Israel Aug 31 '12 at 21:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.