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I want to solve the following differential equation: $$ 2f'(x)(2x+1)+\frac{\kappa}{2}f"(x)x(x+1)=f(x)(\frac{-2b}{x+1}+\frac{2c}{x}+2a) $$

where $\kappa,a,b,c$ are arbitrary positive constants.

Is there anyway to do this? Thanks.

Iew
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1 Answers1

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Your equation is of 2nd order and has only three regular singular points on $\mathbb{P}^1$: namely, $x=0$, $-1$ and $\infty$. In other words, this is a particular case of the Riemann's differential equation.

Therefore, by making the substitution $f(x)=x^{\alpha}(1+x)^{\beta}g(-x)$ and suitably choosing $\alpha$ and $\beta$, it can certainly be reduced to the Gauss hypergeometric equation for $g(t)$ and solved in terms of the hypergeometric functions $_2F_1$.

Start wearing purple
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  • I came across the statement that you formulated above several times. However I have no idea what substitution shall I using to do the reduction. Consider an equation: \begin{equation} f^{''}(x) + \left(\frac{A}{x^{\beta_1}} + \frac{B}{x^{\beta_2}}\right) f(x) = 0 \end{equation} where $\beta_1 >0$ and $\beta_2>0$. This equation has a singular point at zero( Am I right??). What kind of substitution shall I be using to transform it to the hypergeometric equation then? (http://math.stackexchange.com/questions/1018897/an-ordinary-differential-equation-with-time-varying-coefficients) – Przemo Jan 14 '15 at 10:55
  • @Przemo Here the coefficients are polynomials (or rational functions) so the situation is different from yours. In your case, the only substitution I can think of is $r_t=t^{\gamma_1} g(t^{\gamma_2})$. One should first check for which values of $\gamma_{1,2}$ the equation for $g(s)$ will have coefficients given by rational functions of $s$, as only for these values one can hope to reduce the problem to hypergeometric equation or one of its degenerations. Maybe the general case does not admit such a reduction. – Start wearing purple Jan 15 '15 at 12:42