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The hypergeometric equation

$$x(1-x)y''+(2-(1+a+b)x)y'-aby=0$$

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

a,b constant, has a solution of the form $$y=1+\sum_{n=1}^{\infty }a_nx^n$$

despite the fact that $x=0$ is a singular point. Find this solution (i.e. the coefficients $a_n$).

$$y'=\sum_{n=1}^{\infty }a_nnx^{n-1}$$$$y''=\sum_{n=1}^{\infty }a_nn(n-1)x^{n-2}$$

But where to from here?

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Hypergeometric equation ‎$$x(1-x)y^{\prime\prime}+[2-(a+b+1)x]y^\prime-aby=0$$‎ as $1-c=1-2\notin{\Bbb Z}-{\Bbb N}$, let the answer is $$y=x^0(a_0+a_1x+a_2x^2+a_3x^3+\cdots)=a_0+\sum_{n=1}^{\infty }a_nx^n$$‎ with substitution in equation \begin{eqnarray*}‎ && x(1-x)y''+(2-(1+a+b)x)y'-aby= 0\\‎ && x(1-x)\sum_{n=1}^{\infty }a_nn(n-1)x^{n-2}+(2-(1+a+b)x)\sum_{n=1}^{\infty }a_nnx^{n-1}-ab-ab\sum_{n=1}^{\infty }a_nx^n= 0\\‎ && \sum_{n=1}^{\infty }a_nn(n-1)x^{n-1}-\sum_{n=1}^{\infty }a_nn(n-1)x^{n}+2\sum_{n=1}^{\infty }a_nnx^{n-1}-(1+a+b)\sum_{n=1}^{\infty }a_nnx^{n}-ab-ab\sum_{n=1}^{\infty }a_nx^n= 0\\‎ && \sum_{n=2}^{\infty }a_nn(n-1)x^{n-1}-\sum_{n=2}^{\infty }a_nn(n-1)x^{n}+2\sum_{n=1}^{\infty }a_nnx^{n-1}-(1+a+b)\sum_{n=1}^{\infty }a_nnx^{n}-ab-ab\sum_{n=1}^{\infty }a_nx^n=0\\‎ && (2a_1-ab)+x(6a_2+a_1(1+a+b-ab))x+\cdots ‎\end{eqnarray*}‎ $$a_1=\frac{ab}{2\times1}~~~,~~~a_2=\frac{a(a+1)b(b+1)}{2(2+1)1\times2}~~~,~~~a_3=\frac{a(a+1)(a+2)b(b+1)(b+2)}{2(2+1)(2+2)1\times2\times3}~~~,~~~\cdots$$‎ and in general ‎$$a_{n+1}=\frac{(a+n)(b+n)}{(2+n)(1+n)}a_n$$‎ ‎then the answer is ‎$$y=1+\frac{ab}{2\times1}x+\frac{a(a+1)b(b+1)}{2(2+1)\times2!}x^2+\frac{a(a+1)(a+2)b(b+1)(b+2)}{2(2+1)(2+2)\times3!}x^3+\cdots$$‎ or ‎$$y=1+\sum_{n=1}^\infty\frac{a(a+1)\cdots(a+n-1)b(b+1)\cdots(b+n-1)}{2(2+1)\cdots(2+n-1)n!}x^n$$‎

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Note that the coefficient of $x^n$ in $(1-x)x y'' = x y'' - x^2 y''$, $(2 - (1+a+b) x) y'$ and $-ab y$ can all be written in terms of $a_n$ and $a_{n+1}$. This gives you a recurrence relation.

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Hint

Let us expand the equation $$x(1-x)y''+(2-(1+a+b)x)y'-aby=0$$

$$x y''-x^2y''+2y' -(1+a+b)xy'-aby=0$$ $$y=1+\sum_{n=1}^{\infty }k_nx^n\implies y'=\sum_{n=1}^{\infty }k_n n x^{n-1}\implies y''=\sum_{n=1}^{\infty }k_n n(n-1) x^{n-2}$$ and replace to get $$\sum_{n=1}^{\infty }k_n n(n-1) x^{n-2}-\sum_{n=1}^{\infty }k_n n(n-1) x^{n}+2\sum_{n=1}^{\infty }k_n n x^{n-1}-(1+a+b)\sum_{n=1}^{\infty }k_n n x^{n}-ab-\sum_{n=1}^{\infty }k_nx^n=0$$ Now, for a given degree $m$, we then have $$k_{m+2}(m+2)(m+1)-k_mm(m-1)+2k_{m+1}(m+1)-(1+a+b)k_mm-k_m=0$$ Simplify to get the recurrence relation.

Concerning the first terms, for sure you will get $$2k_1-ab=0$$ $$6 k_2-(a+1) (b+1) k_1=0$$

I am sure that you can easily take it from here.

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