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I had the differential equation $y'+y=\frac1x$, which I solved for $y$ as a power series:

$$y=\frac1x\sum_{n=0}^{\infty}\frac{n!}{x^n}$$

Which was a power series at $\infty$, so it doesn't really help me much.

So my first question is whether or not $y$ is solvable here (as a power series if needed) where it actually converges.

My second question is if it pure coincidence that the summation is very similar to the power series of $e^x$.

$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$

Is there some reason for their very similar forms, or just my stumbling upon these two unrelated power series?

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Multiplying both sides of the equation by $e^x$ and reversing the product rule gives $$(e^x y)' = \frac{e^x}{x} .$$

The r.h.s. famously does not admit a closed-form antiderivative, but it can be written in terms of the exponential integral function. Integrating both sides and rearranging gives (for $x > 0$) the general solution $$y = e^{-x} \left( \int_1^x \frac{e^t}{t} \,dt + C \right) .$$

One can certainly expand this in a power series (about some $x_0 > 0$), but probably it's not very illuminating.

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Clear that the homogeneous solution is $y=Ae^{-x}$ so the final solution should be $$y = Ae^{-x} + f(x)$$ where $f$ is a particular solution.

Wolfram Alpha suggests the solution using special functions, so one does exist, but is not expressible in elementary terms.

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  • $\begingroup$ So we can't solve for $y$ even with a power series (that converges)? $\endgroup$ – Simply Beautiful Art Apr 11 '16 at 22:38

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