# Solve $x(1-x)y''+2(1-2x)y'-2y=0$ by the Frobenius Method

Find the second solution. First solution is $$\dfrac 1 {1-x}$$. Solve by Frobenius Method: $$x(1-x)y''+2(1-2x)y'-2y=0\,.$$

The first solution I am able to get is $$\dfrac{1}{1-x}$$. Other solution is $$\dfrac{1}{x}$$, but I am getting $$-\dfrac{1}{x(1-x)}$$. Where am I going wrong?

• What have you tried explicitly? See How to ask a good question. – Saad May 1 at 13:43
• To prevent further close votes, I would like to state that the OP actually solved the problem completely, but did not realize it. Readers do not even have to try to solve the problem to see that the OP did finish the problem. If this doesn't indicate effort, I don't know what will. Please see the first part of my answer below. – Batominovski May 1 at 15:05
• @Batominovski the question is not only if it indicates effort but if it is presented well enough. The presentation is overly terse. – quid May 1 at 15:43
• @quid I think we are moving goal posts if we keep adding criteria for what constitute as good questions. I think the OP of this question did well enough. I didn't even have to add a full solution to answer the OP's question. But again, I may be less strict. – Batominovski May 1 at 15:45
• @Batominovski I will continue the exchange on meta and not here. – quid May 1 at 15:57

As for your attempt, you did nothing wrong. Note that $$\frac{1}{x}=(-1)\left(\frac{1}{1-x}\right)+(-1)\left(-\frac{1}{x(1-x)}\right)\,.$$ That is, you can take the two linearly independent solutions of your differential equation to be $$y(x)=\frac1{1-x}\text{ and }y(x)=\frac{1}{x}\,,$$ but you can take them to be $$y(x)=\frac1{1-x}\text{ and }y(x)=-\frac{1}{x(1-x)}$$ as well.
Here is a full solution as requested, although this seems unnecessary, because you have actually solved the problem. Suppose that $$y:U\to\mathbb{R}$$ is a twice differentiable function on an open set $$U\subseteq \mathbb{R}$$ which satisfies $$x(1-x)\,y''(x)+2(1-2x)\,y'(x)-2\,y(x)=0$$ for all $$x\in U$$. We make an Ansatz that $$y(x)=\sum_{n=0}^\infty a_nx^{n+r}$$ where $$r$$, $$a_0$$, $$a_1$$, $$a_2$$, $$\ldots$$ are real numbers.
We have $$y'(x)=\sum_{n=0}^\infty\,(n+r)\,a_n\,x^{n+r-1}\,,$$ and $$y''(x)=\sum_{n=0}^\infty\,(n+r)(n+r-1)\,a_n\,x^{n+r-2}\,.$$ Thus, $$x(1-x)\,y''(x)=\small r(r-1)\,a_0\,x^{r-1}+\sum_{n=0}^\infty\,\big((n+r+1)(n+r)\,a_{n+1}-(n+r)(n+r-1)\,a_n\big)\,x^{n+r}\,,$$ and $$2(1-2x)\,y'(x)=2r\,a_0\,x^{r-1}+\sum_{n=0}^\infty\,\big(2(n+r+1)\,a_{n+1}-4(n+r)\,a_n\big)\,x^{n+r-1}\,.$$ Ergo, \begin{align}0&=x(1-x)\,y''(x)+2(1-2x)\,y'(x)-2\,y(x) \\&=r(r+1)\,a_0\,x^{r-1}+\sum_{n=0}^\infty\,(n+r+1)(n+r+2)\,\big(a_{n+1}-a_n\big)\,x^{n+r}\,. \end{align} It follows immediately that $$r(r+1)=0\text{ and }(n+r+1)(n+r+2)\,\big(a_{n+1}-a_n\big)=0$$ for $$n=0,1,2,\ldots$$.
If $$r=0$$, then $$(n+1)(n+2)\,(a_{n+1}-a_n\big)=0$$ for all $$n=0,1,2,\ldots$$. Thus, $$a_n=a_0$$ for every $$n=0,1,2,\ldots$$, and we may write $$y(x)=a_0\,y_1(x)\,,\text{ where }y_1(x):=\sum_{n=0}^\infty\,x^{n+0}=\frac{a_0}{1-x}\,.$$ The power series solution is valid only for $$U=(-1,+1)$$, but $$y_1(x)=\dfrac{1}{1-x}$$ works even if $$U=\mathbb{R}\setminus\{1\}$$.
If $$r=-1$$, then $$n(n+1)\,\big(a_{n+1}-a_n\big)=0$$ for all $$n=0,1,2,\ldots$$. Thus, $$a_n=a_1$$ for every $$n=1,2,3,\ldots$$. We may assume that $$a_1=0$$ (otherwise, we may subtract an appropriate multiple of $$y_1(x)$$ out of $$y(x)$$). This means $$y(x)=a_0\,y_2(x)\,,\text{ where }y_2(x):=\frac{1}{x}\,.$$ Note that we may take $$U$$ to be $$\mathbb{R}\setminus\{0\}$$ in this case.
The general solutions are $$y(x)=\frac{A}{1-x}+\frac{B}{x}$$ which are defined on $$U=\mathbb{R}\setminus\{0,1\}$$, although $$A$$ and $$B$$ must be treated as local constants (i.e., they are constant on three different intervals: $$(-\infty,0)$$, $$(0,1)$$, and $$(1,\infty)$$).