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I am trying to solve a equation which I have already solved using oDE but I want to solve it using a power series expansion but how do I express y as a power series?

Equation is as below

$$ (1+x) \, y'(x)=y(x) $$

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  • $\begingroup$ You write $y$ as power series with unknown coefficients and insert it into the ODE to get conditions on the unknowns. Then you try to solve this system to determine the unknown coefficients. $\endgroup$
    – mvw
    Nov 23 '15 at 13:24
  • $\begingroup$ Sorry my question was more referring to how I write y as a power series $\endgroup$
    – darren
    Nov 23 '15 at 13:28
  • $\begingroup$ The question is not precise, you have to specify around which point you would like to have a solution. The answer strongly depend on this data. $\endgroup$
    – Thomas
    Nov 23 '15 at 13:30
  • $\begingroup$ Maybe you should have a look here first: en.wikipedia.org/wiki/Power_series $\endgroup$
    – mvw
    Nov 23 '15 at 13:32
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If you search online, you will find many sources and examples on the web. such as:

Link 1

Link 2

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let $(a_n)$ so $$ y(x)=\sum_0^\infty a_nx^n$$ then $$y'(x)=\sum_1^\infty a_nn(x-a)^{n-1}=\sum_0^\infty a_{n+1}(n+1)x^{n}$$ $$(1+x)y'(x)=y(x) \iff (1+x)\sum_0^\infty a_{n+1}(n+1)x^{n}=\sum_0^\infty a_nx^n$$ $$\iff \sum_0^\infty a_{n+1}(n+1)x^{n}+\sum_0^\infty a_{n+1}(n+1)x^{n+1}=\sum_0^\infty a_nx^n$$ $$\iff \sum_0^\infty a_{n+1}(n+1)x^{n}+\sum_1^\infty a_{n}nx^{n}=\sum_0^\infty a_nx^n$$ $$\iff a_1+\sum_1^\infty (a_{n+1}(n+1) + a_{n}n) x^{n}=\sum_0^\infty a_nx^n$$ $$\iff a_1=a_0 \; and \; \forall n \ge 1, a_{n+1}(n+1)=-a_n(n-1)$$ $$\iff a_1=a_0 \; and \; \forall n > 1, a_n=0$$

So $\forall f$ solution of the equation $\exists A\in\mathbb{R}$ / f(x)=A(x+1) and reciprocally , all these functions are solutions

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