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How can I solve the following equation:

$$z^2y''(z)+\frac{1}{\gamma}zy'(z)(\gamma-a_1-a_2 z)+\frac{1}{\gamma^2}y(z)(b_1+b_2z+b_3 z^2)=0$$

Apparently I am supposed to get to a confluent hypergeometric equation but I don't know how.

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    $\begingroup$ have you missed a y out of the last term? $\endgroup$ – user121049 Apr 11 '17 at 9:02
  • $\begingroup$ Yes, I fixed it $\endgroup$ – NSZ Apr 11 '17 at 9:23
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HINT : Change of variable and function $$\begin{cases} z=\alpha x \\ y(z)=x^\beta e^{\mu x}u(x) \end{cases}$$ In the transformed ODE, determine the parameters $\alpha$ , $\beta$ , $\mu$ to simplify it to the confluent hypergeometric form: $$xu''(x)+(C-x)u'(x)-Au(x)=0$$ This will lead to the expressions of $\alpha$ , $\beta$ , $\mu$ , $A$ and $C$ as functions of the parameters present in the original ODE. Boring calculus, perseverance and good luck !

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