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Consider $$ x^2 y''+y'=x-1, $$ where $y=y(x)$ and $x>0$. I've tried the power method but obtained a divergent series. Any help would be appreciated.

Update

The equation can be rewritten as $$ y''+\frac{1}{x^2}y'=\frac{x-1}{x^2}, $$ and then first solved for $z=y'$ using the integrating factor approach. The solution is $$ z=y'=e^{1/x}\int^x e^{-1/t}\frac{t-1}{t^2} dt, $$ whence $$ y=\int^x\left[e^{1/s}\int^s e^{-1/t}\frac{t-1}{t^2} dt\right]ds. $$

Is there a way to express this solution in terms of elementary functions with no integration involved? I just fail to see one.

I would actually be interested in finding an expansion of the solution into a series of powers of $x$. How would I go about obtaining such an expansion?

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  • $\begingroup$ Have you tried integrating factor? $\endgroup$ – Arian Feb 27 '15 at 12:56
  • $\begingroup$ Yes, of course, that did give the solution in a closed form (as a double integral that cannot be evaluated), but I'd like to have the solution as a series. $\endgroup$ – Jason Feb 27 '15 at 12:59
  • $\begingroup$ i think a solution in the known elementary function is not possible $\endgroup$ – Dr. Sonnhard Graubner Feb 27 '15 at 13:05
  • $\begingroup$ It's a first order with $z=y'$ $\endgroup$ – Maman Feb 27 '15 at 13:15
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    $\begingroup$ In general (as in this case), for a linear differential equation at an irregular singular point, series solutions have radius of convergence zero. $\endgroup$ – GEdgar Feb 27 '15 at 16:47

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