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!
 A: $$\sum_i(4i(i-1)b_i+2ib_i )x^{i-1}+(-2ib_i-b_i)x^i=\sum_i(2(i+1)b_{i+1}-b_i)(2i+1)x^i$$
A: 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 
$$y=\sum_{i=0}^\infty{\frac{x^i}{2^ii!}}=\sum_{i=0}^\infty{\frac{(\frac{x}{2})^i}{i!}}=e^{\frac{x}{2}}$$
