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Find a power series solution for the first order D.E.: $$(1-x)y'-2y=0$$$$\sum a_nx^n, y(0)=1$$$$y'=\sum_{n=0}^{\infty }na_nx^{n-1}$$$$(1-x)y'=y'-xy'$$$$xy'=x\sum_{n=0}^{\infty }na_nx^{n-1}=\sum_{n=0}^{\infty }na_nx^{n}$$$$\sum_{n=0}^{\infty }na_{n+1}x^{n}-\sum_{n=0}^{\infty }na_nx^{n}-\sum_{n=0}^{\infty }2a_nx^{n}=0$$$$a_0=y(0)=1$$$$x^n\begin{bmatrix} \sum_{n=0}^{\infty }na_{n+1}-\sum_{n=0}^{\infty }na_n-\sum_{n=0}^{\infty }2a_n \end{bmatrix}=0$$ But where to from here? If $x\neq0$ $$\begin{bmatrix} \sum_{n=0}^{\infty }na_{n+1}-\sum_{n=0}^{\infty }na_n-\sum_{n=0}^{\infty }2a_n \end{bmatrix}=0$$

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    $\begingroup$ Watch out when you're taking the derivative of a series. The series for $y'$ should start at $n = 1$ $\endgroup$ – Kaynex Aug 2 '17 at 2:34
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$$(1-x)\frac{dy}{dx}-2y=0$$

We assume that a power series solution exists for this differential equation

$$y=\sum_{n=0}^{\infty}c_{n}x^{n}$$

Differentiating we got,

$$\frac{dy}{dx}=\sum_{n=1}^{\infty}nc_{n}x^{n-1}$$

Now substituting it into the differential equation,

$$(1-x)\sum_{n=1}^{\infty}nc_{n}x^{n-1}-2\sum_{n=0}^{\infty}c_{n}x^{n}=0$$

$$\sum_{n=1}^{\infty}nc_{n}x^{n-1}-\sum_{n=1}^{\infty}nc_{n}x^{n}-2\sum_{n=0}^{\infty}c_{n}x^{n}=0$$

Making index shift,

$$\sum_{n=0}^{\infty}(n+1)c_{n+1}x^{n}-\sum_{n=1}^{\infty}nc_{n}x^{n}-2\sum_{n=0}^{\infty}c_{n}x^{n}=0$$

Taking out a few terms,

$$c_1-2c_0+\sum_{n=1}^{\infty}(n+1)c_{n+1}x^{n}-\sum_{n=1}^{\infty}nc_{n}x^{n}-2\sum_{n=1}^{\infty}c_{n}x^{n}=0$$

Assuming $c_0\neq 0$

$$c_0=\frac{c_1}{2}$$

The recurrence formula is then,

$$c_{n+1}=\frac{(n+2)c_n}{n+1}, n\geq 1$$

We can obtain our first few terms,

$$c_{2}=\frac{3c_1}{2}$$

I hope that you can move on.

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    $\begingroup$ $$c_{n+1}=\frac{n+2}{n+1}c_{n}=\frac{n+2}{n+1}\frac{n+1}{n}c_{n-1}=\frac{n+2}{n+1}\frac{n+1}{n}\frac{n}{n-1}c_{n-2}=.....=\frac{(n+2)!}{(n+1)!}c_{0}$$ Thus $$c_{n}=\frac{(n+1)!}{n!}c_{0}$$ and the solution is $$y(x)=c_{0}\sum_{n=0}^{\infty}\frac{(n+1)!}{n!}x^{n}$$ $\endgroup$ – Kiryl Pesotski Aug 3 '17 at 10:40
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    $\begingroup$ And this series, of course, is nothing but a binomial expansion of the solution $$y(x)=\frac{c_{0}}{(1-x)^{2}}$$ $\endgroup$ – Kiryl Pesotski Aug 3 '17 at 10:41

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