Find solutions of the given equation in the form of power series Find solutions of the given equation in the form of power series
$$y'' +2xy' = 0$$
My approach:
$$2xy' = 2a_1x + 2\times2a_2x^2 + 2\times3a_3x^3 + ... = \sum 2ia_{i}x^{i}$$
$$y'' = 2a_2 + 3\times2a_3x + 4\times3a_4x^2 + ... = \sum (i+1)(i+2)a_{i+2}x^{i}$$ 
So to have $y'' +2xy' = 0$ we need to have every coefficient equal to $0$.
So
$x^0$ // $2a_2 + 0 = 0 => a_2=0$
$x^1$ // $3\times2a_3 + 2a_1 = 0 => a_3 = -a_1/3$
$x^2$ // $4\times3a_4 + 2\times2a_2=0 => a_4 = -4a_2/12 = 0$
...
So it was easy to observe that every even coefficient is equal to $0$ and every odd one can be written in terms of $a_0$
Afterwards I wanted to find a recursive formula for odd coefficients and this is a problem that I can't solve. 
I tried doing it in that way:
$$2n(2n+1)a_{2n+1} + 2(2n-1)a_{2n-1} = 0$$
$$a_{2n+1} = -\frac{2(2n-1)a_{2n-1}}{2n(2n+1)}$$
Now probably it should be somehow easy to write it as a recursive formula in terms of $a_0$, but I can't really see it. 

Let the solution of the given ODE be $y=\sum c_{s}x^{s}$.
Then this satisfies the equation $y^{''}+2xy^{'}=0$.
Now, $y^{'}=\sum sc_{s}x^{s-1}$ and $y^{''}=\sum s(s-1)c_{s}x^{s-2}$. Substituting these values in the differential equation.
 $\sum s(s-1)c_{s}x^{s-2} + 2x \sum sc_{s}x^{s-1}=0$
$=>\sum s(s-1)c_{s}x^{s-2} + 2\sum sc_{s}x^{s}=0$
$=>\sum 0(0-1)c_{0}x^{0} + \sum 1(1-1) c_{1}x^{1} + \sum_{s=2} s(s-1)c_{s}x^{s} + 2\sum 0c_{0}x^{0} + 2\sum_{s=1} sc_{s}x^{s}=0$
$=>\sum_{s=0} (s+1)(s+2)c_{s+2}x^{s} + 2\sum_{s=0} (s+1)c_{s+1}x^{s+1} =0$

Now since the last relation is an identity, therefore
$(s+1)(s+2)c_{s+2}=0$ and $2(s+1)c_{s+1}=0$

$=>(s+1)(s+2)c_{s+2}=2(s+1)c_{s+1}$

$=>c_{s+2}=\frac{2c_{s+1}}{s+2}$

$=>c_{s+1}=\frac{2c_{s}}{s+1}$

Now I think you can see the recurrence relation
