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So, Im trying to self-learn method of frobenius, and I would like to ask if someone can explain to me how can we solve the following DE about $ x = 0$ using this method.

$$ x(1 - x)y'' - 3xy' - y = 0 $$

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The motivation behind Frobenius method is to seek a power series solution to ordinary differential equations.

Let $y(x) = \displaystyle \sum_{n=0}^{\infty} a_n x^n$. Then we get that $$y'(x) = \sum_{n=0}^{\infty} na_n x^{n-1}$$ $$3xy'(x) = \sum_{n=0}^{\infty} 3na_n x^{n}$$ $$y''(x) = \sum_{n=0}^{\infty} n(n-1)a_n x^{n-2}$$ $$xy''(x) = \sum_{n=0}^{\infty} n(n-1)a_n x^{n-1} = \sum_{n=0}^{\infty} n(n+1)a_{n+1} x^{n}$$ $$x^2y''(x) = \sum_{n=0}^{\infty} n(n-1)a_n x^{n}$$ The ODE is $$xy'' - x^2 y'' -3xy' - y = 0$$ Plugging in the appropriate series expansions, we get that $$\sum_{n=0}^{\infty} \left(n(n+1)a_{n+1} - n(n-1)a_n - 3na_n - a_n\right)x^n = 0$$ Hence, we get that $$n(n+1)a_{n+1} = (n(n-1) +3n+1)a_n = (n+1)^2 a_n \implies a_{n+1} = \dfrac{n+1}{n}a_n$$ First note that $a_0 = 0$. Choose $a_1$ arbitrarily. Then we get that $a_2 = 2a_1$, $a_3 = 3a_1$, $a_4 = 4a_1$ and in general, $a_{n} = na_1$. Hence, the solution is given by $$y(x) = a_1 \left(x+2x^2 + 3x^3 + \cdots\right)$$ This power series is valid only within $\vert x \vert <1$. In this region, we can simplify the power series to get \begin{align} y(x) & = a_1 x \left(1 + 2x + 3x^2 + \cdots \right)\\ & = a_1 x \dfrac{d}{dx} \left(x + x^2 + x^3 + \cdots \right)\\ & = a_1 x \dfrac{d}{dx} \left(\dfrac{x}{1-x}\right)\\ & = a_1 \dfrac{x}{(1-x)^2} \end{align}

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Thanks a lot!!! –  Learner Oct 28 '12 at 5:35

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