解を
\[ y\left(x\right)=\sum_{k=0}^{\infty}u_{k}x^{k+h} \] と仮定するし、\(u_{0}\ne0\)とする。
これを超幾何微分方程式に代入すると、
\begin{align*} 0 & =x\left(1-x\right)y''\left(x\right)+\left(c-\left(a+b+1\right)x\right)y'\left(x\right)-aby\left(x\right)\\ & =x\left(1-x\right)\sum_{k=0}^{\infty}\left(k+h\right)\left(k+h-1\right)u_{k}x^{k+h-2}+\left(c-\left(a+b+1\right)x\right)\sum_{k=0}^{\infty}\left(k+h\right)u_{k}x^{k+h-1}-ab\sum_{k=0}^{\infty}u_{k}x^{k+h}\\ & =\sum_{k=0}^{\infty}\left(k+h\right)\left(k+h-1\right)u_{k}x^{k+h-1}-\sum_{k=0}^{\infty}\left(k+h\right)\left(k+h-1\right)u_{k}x^{k+h}+c\sum_{k=0}^{\infty}\left(k+h\right)u_{k}x^{k+h-1}-\left(a+b+1\right)\sum_{k=0}^{\infty}\left(k+h\right)u_{k}x^{k+h}-ab\sum_{k=0}^{\infty}u_{k}x^{k+h}\\ & =\sum_{k=0}^{\infty}\left(k+h\right)\left(k+h+c-1\right)u_{k}x^{k+h-1}-\sum_{k=0}^{\infty}\left(\left(k+h\right)\left(k+h+a+b\right)+ab\right)u_{k}x^{k+h}\\ & =\sum_{k=-1}^{\infty}\left(k+h+1\right)\left(k+h+c\right)u_{k+1}x^{k+h}-\sum_{k=0}^{\infty}\left(\left(k+h\right)^{2}+\left(a+b\right)\left(k+h\right)+ab\right)u_{k}x^{k+h}\\ & =h\left(h+c-1\right)u_{0}x^{h-1}+\sum_{k=0}^{\infty}\left(k+h+1\right)\left(k+h+c\right)u_{k+1}x^{k+h}-\sum_{k=0}^{\infty}\left(k+h+a\right)\left(k+h+b\right)u_{k}x^{k+h}\\ & =h\left(h+c-1\right)u_{0}x^{h-1}+\sum_{k=0}^{\infty}\left\{ \left(k+h+1\right)\left(k+h+c\right)u_{k+1}-\left(k+h+a\right)\left(k+h+b\right)u_{k}\right\} x^{k+h} \end{align*} これが成り立つには
\[ \begin{cases} h\left(h+c-1\right)=0\\ u_{k+1}=\frac{\left(k+h+a\right)\left(k+h+b\right)}{\left(k+h+1\right)\left(k+h+c\right)}u_{k} \end{cases} \] これより、\(h=0\lor h=1-c\)となり、
\begin{align*} u_{n} & =u_{0}\prod_{k=0}^{n-1}\frac{u_{k+1}}{u_{k}}\\ & =u_{0}\prod_{k=0}^{n-1}\frac{\left(k+h+a\right)\left(k+h+b\right)}{\left(k+h+1\right)\left(k+h+c\right)}\\ & =u_{0}\frac{Q\left(a+h,n\right)Q\left(b+h,n\right)}{Q\left(c+h,n\right)Q\left(1+h,n\right)} \end{align*} \begin{align*} y\left(x\right) & =\sum_{k=0}^{\infty}u_{k}x^{k+h}\\ & =u_{0}\sum_{k=0}^{\infty}\frac{Q\left(a+h,k\right)Q\left(b+h,k\right)}{Q\left(c+h,k\right)Q\left(1+h,k\right)}x^{k+h}\\ & =u_{0}x^{h}\sum_{k=0}^{\infty}\frac{Q\left(a+h,k\right)Q\left(b+h,k\right)Q\left(1,k\right)}{Q\left(c+h,k\right)Q\left(1+h,k\right)}\frac{x^{k}}{k!}\\ & =u_{0}x^{h}F\left(a+h,b+h,1;c+h,1+h;x\right) \end{align*} これより、\(h=0\)のとき\(u_{0}=1\)とすると、
\begin{align*} y\left(x\right) & =F\left(a,b,1;c,1;x\right)\\ & =F\left(a,b;c;x\right) \end{align*} \(h=1-c\)のとき\(u_{0}=1\)とすると、
\begin{align*} y\left(x\right) & =x^{1-c}F\left(a-c+1,b-c+1,1;1,2-c;x\right)\\ & =x^{1-c}F\left(a-c+1,b-c+1;2-c;x\right) \end{align*} となる。
従って解は
\[ \begin{cases} y\left(x\right)=F\left(a,b;c;x\right)\\ y\left(x\right)=x^{1-c}F\left(a-c+1,b-c+1;2-c;x\right) \end{cases} \] となり題意は成り立つ。