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I want to solve the differential equation $f'(x)=f(x)$ using power series of the form $$f(x)=\sum_{n=0}^{\infty}{c_nx^n}$$

From my previous knowledge I know that the solution is $f(x)=c_0e^x$ I can justify that using MacLaurin series, but how do I show the process? Also, why does $c_n=c_{n-1}$?

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    $\begingroup$ Hint: What is the power series of $f'$? $\endgroup$
    – sranthrop
    Commented Nov 17, 2015 at 4:20

3 Answers 3

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It is not true that $c_n = c_{n-1}$, you must have made a mistake in your differentiation.

The actual recurrence relation you should obtain for the coefficients is $c_n = c_{n-1} / n$. This equation is easy to solve, and it indeed does give the coefficients for $e^x$ as you expect it to.

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Hint: $$ \sum_{n=0}^\infty c_nx^n = \frac d{dx} \sum_{n=0}^\infty c_nx^n = \sum_{n=0}^\infty c_n \frac d{dx}x^n $$

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Heuristics: The problem can be solved once you notice that $$ De^{x} = D\bigg( 1 + x + \frac{x^{2}}{2!} + \cdots \bigg) = 0 + 1 + x + \frac{x^{2}}{2!} + \cdots = e^{x} $$ for all $x \in \Bbb{R}$; there is no "missing" term after differentiation.

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