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!

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 
$$y=\sum_{i=0}^\infty{\frac{x^i}{2^ii!}}=\sum_{i=0}^\infty{\frac{(\frac{x}{2})^i}{i!}}=e^{\frac{x}{2}}$$
