I Derivation for the indicial exponent in the Frobenius method

Kloo
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I'm reading a book called Asymptotic Methods and Perturbation Theory, and I came across a derivation that I just couldn't follow. Maybe its simple and I am missing something. Equation 3.3.3b below. y(x) takes the form A(x)*(x-x0)^α and A(x) is expanded in a taylor series.

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Use \begin{split}<br /> \left(\sum_{n=0}^\infty a_nx^n \right)\left(\sum_{n=0}^\infty b_nx^n \right) &amp;= <br /> \sum_{n=0}^\infty \sum_{m=0}^\infty a_n b_m x^{n+m} \\<br /> &amp;=\sum_{k=0}^\infty \left(\sum_{n=0}^n a_n b_{k-n}\right) x^k. \end{split}
 
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I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
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