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For the ODE: $$\frac{dy}{dx}=2y$$

If the successive derivatives calculated are:

$$y'=2y,y''=2y'=(2^2)y,y^{(3)}=(2^3)y,\ldots,y^{(n)}=(2^n)y$$

How do I find the coefficients of the following equation for an infinite series? (if the coefficients are $a_1, a_2, a_3, \ldots$)

$$y(x)=a_0 +a_1x^1+a_2x^2 +a_3x^3 + \ldots +a_nx^n + \ldots$$

Also, How could you write $y(x)$ as an infinite series? Can I use Maclaurin Series?

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You absolutely want to use Maclaurin series. Let me help you remember how that goes.

If $f(x)$ is a sufficiently nice function then $$ f(x) = f(a) + f'(a)(x-a) + \frac12 f''(a)(x-a)^2 + \cdots + \frac{1}{n!}f^{(n)}(a)(x-a)^n + \cdots. $$

For your particular problem, let's say that $f(x)$ is solution to the ODE, and let's pretend that we know $f(0)$. Call it $y_0$. Then $f'(0)=2f(0)=2y_0$. And so on -- you can figure out $f^{(n)}(0)$ in terms of $y_0$ using the pattern you found.

Note: I explained your example using a Taylor series, which is a Maclaurin series with $a=0$. I told you about Maclaurin series because, for example, you could use them to solve your problem if I told you what $f(1)$ was instead of $f(0)$.

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  • $\begingroup$ what you have is called the taylor series; not maclaurin series. $\endgroup$ – abel May 17 '15 at 23:52

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