Alternative solution for hypergeometric equation

Question

I need to get another solution for

$$ x(1-x) \frac{\partial^{2}}{\partial x^2} y(x) + (c-(a+b+1)x)\frac{\partial}{\partial x} y(x) - aby(x) = 0$$

by substituting with $y=x^{1-c}z$. When I do the substitution I get really long derivatives there, but in the end the answer should look like

$$ x(1-x) \frac{\partial^{2}}{\partial x^2} z(x) + ((2-c)-((a-c+1)+(b-c+1)+1)x)\frac{\partial}{\partial x} z(x) - (a-c+1)(b-c+1)z(x) = 0$$

So there is no $x^{1-c}$ in this equation which leads me to think that I am doing something wrong.

Thanks for any hints

Answer 1

You did not differentiate properly $y=z(x)x^{1-c}$

You should get $y'=x^{1-c} z'(x)+(1-c) x^{-c} z(x)$

and $y''=x^{1-c} z''(x)+2 (1-c) x^{-c} z'(x)-(1-c) c x^{-c-1} z(x)$