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How can I solve this ODE:


Can you please also show the derivation.

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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
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2 Answers

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)

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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.

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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
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