By finding solutions as power series in $x$ solve $4xy''+2(1-x)y'-y=0 .$

By finding solutions as power series in $x$ solve $$4xy''+2(1-x)y'-y=0 .$$ What I did is the following. First I let the solution $y$ be equal to $$y =\sum_{i=0}^{\infty} b_ix^i =b_0 +b_1x+b_2x^2+\ldots$$ for undetermined $b_i$. Then I found the expression for $y'$ and $y''$, $$y' =\sum_{i=0}^{\infty} ib_ix^{i-1} =b_1 + 2b_2x+3b_3x^2+\ldots.$$ and $$y'' =\sum_{i=0}^{\infty} i(i-1)b_ix^{i-2} =2b_2+6b_3x+12b_4x^2\ldots.$$ Now I put these in the original DE to get $$4\sum i(i-1)b_ix^{i-1}+2\sum ib_i(x^{i-1}-x^i) - \sum b_ix^i =0$$ where all sums range from $0$ to infinity. Finally this becomes $$\sum \left\{ (4i(i-i)b_i+2ib_i )x^{i-1}+(-2ib_i-b_i)x^i \right\}=0.$$ At this point I am fairly certain I have already made a mistake somewhere, probably in working out the power series of $y'$ or $y''$. Who can help point it out to me, I am pretty sure in the last sum there should be terms like $b_{i+1}$ or $b_{i+2}$. Thanks for any help or tips!

EDIT I have gotten further by realizing that $$y' =\sum_{i=0}^{\infty} ib_ix^{i-1} =\sum_{i=1}^{\infty} ib_ix^{i-1}=\sum_{i=0}^{\infty} (i+1)b_{i+1}x^{i}$$ and $$y'' =\sum_{i=0}^{\infty} (i+2)(i+1)b_{i+2}x^{i}.$$ Putting these in the original DE I get $$\sum \left\{ [4(i+2)(i+1)b_{i+2}-2(i+1)b_{i+1}]x^{i+1} + [2(i+1)b_{i+1}-b_i]x^i \right\}=0.$$ This must be true for all $x$ and thus we have $$4(i+2)(i+1)b_{i+2}=2(i+1)b_{i+1}$$ and $$2(i+1)b_{i+1} = b_i.$$ After simplyfying these two conditions are seen to be identical. Now I've set $b_0=1$ to obtain the solution $$y = 1 + \frac{x}{2}+ \frac{x^2}{8} +\frac{x^3}{48}+\ldots + \frac{x^i}{2^i(i!)}+\ldots.$$ Now I've arrived at the ackward position where in working out the question here I have actually managed to solve it. My last question is then, does anyone recognize this power series? Thanks!

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Yes you solved it. Next you might want to post your solution as an answer (this is even recommanded, and is surely better than to accept a flawed answer). About your last power series, surely you would recognize $\sum\limits_i\frac{x^i}{i!}$? Compare to your case... – Did Jan 5 '14 at 13:41
Ah thanks a lot for the tip a the end. I do indeed recognize that series :) so the answer is $e^{\frac{x}{2}}$ or if I had chosen an arbitrary $b_0 =C$ I would have $Ce^{\frac{x}{2}}$, however, shouldn't I have found two independent solutions? – Slugger Jan 5 '14 at 14:12
Indeed, but the other solution is not a series in x, hence one cannot "see" it with this approach. – Did Jan 5 '14 at 14:17

You have made your mistake in the power series. In particular, you need to end up with a recurrence relation and solve that. $$y'=\sum_{i=0}^\infty{ib_ix^{i-1}}=0+b_1+2b_2x+3b_3x^2+...=\sum_{i=1}^\infty{ib_ix^{i-1}}$$ Now you need to get your lower bound so that it starts at $0$. Rewriting the sum using $i=0$, we get that $$\sum_{i=1}^\infty{ib_ix^{i-1}}=\sum_{i=0}^\infty{(i+1)b_{i+1}x^i}$$ Similarly, $$y''=\sum_{i=0}^\infty{i(i-1)b_ix^{i-2}}=0(-1)x^{-2}+1(0)b_1x^{-1}+2(1)b_2+3(2)b_3x+...=\sum_{i=2}^\infty{ib_ix^{i-2}}$$ Now rewrite that also with an index of 0. $$\sum_{i=2}^\infty{i(i-1)b_ix^{i-2}}=\sum_{i=0}^\infty{(i+2)(i+1)b_{i+2}x^i}$$ Since all the indices are now $0$, you can rewrite the equation as
$$4x\sum_{i=0}^\infty{(i+2)(i+1)b_{i+2}x^i}+2(1-x)\sum_{i=0}^\infty{(i+1)b_{i+1}x^i}-\sum_{i=o}^\infty{b_ix^i}=0$$ You now have one more issue to resolve. You have to include the factors of $x$ in both the $y''$ and $y'$ sums and this gives you two higher powers of $x$. You'll again have to rewrite the sums so that each sum contains sums of $x^i$, not $x^{i+1}$.
$$\sum_{i=0}^\infty{4i(i+1)b_{i+1}x^i}+\sum_{i=0}^\infty{2(i+1)b_{i+1}x^i}-\sum_{i=o}^\infty{2ib_ix^i}-\sum_{i=0}^\infty{b_ix^i}$$ $$=\sum_{i=0}^\infty{[2(2i+1)(i+1)b_{i+1}-(2i+1)b_i}]x^i$$ So, we then see that $2(2i+1)(i+1)b_{i+1}=(2i+1)b_i$ and thus $$2(i+1)b_{i+1}=b_i\Rightarrow b_{i+1}=\frac{b_i}{2(i+1)}$$ Setting $b_0=1$, and replacing the $b_i$'s in the series expansion of $y$, we get $$y=\sum_{i=0}^\infty{\frac{x^i}{2^ii!}}=\sum_{i=0}^\infty{\frac{(\frac{x}{2})^i}{i!}}=e^{\frac{x}{2}}$$

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"you can rewrite the equation as one sum which has terms $b_i$, $b_{i+1}$, and $b_{i+2}$" I doubt that. Only consecutive coefficients are involved, say $b_i$ and $b_{i+1}$. – Did Jan 5 '14 at 13:33
Thanks a lot for your answer, I actually recognized my mistake and worked out my answer. Still thanks a lot for your answer! – Slugger Jan 5 '14 at 13:40
Answer still wrong (if that matters). – Did Jan 5 '14 at 13:42
I just put the x into the $y''$ and reduced the $b_{i+2}$ to $b_{i+1}$ so i see where you are leading to, @Did. But is my answer wrong or just not finished? – Eleven-Eleven Jan 5 '14 at 13:46
Ambiguous only in the revised version, since, inserting the prefactors x and 1-x in the series, the coefficient of x^i involves b_i and b_{i+1} only. – Did Jan 5 '14 at 14:02

$$\sum_i(4i(i-1)b_i+2ib_i )x^{i-1}+(-2ib_i-b_i)x^i=\sum_i(2(i+1)b_{i+1}-b_i)(2i+1)x^i$$

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