<s>[INST] $2x^2y''+3xy'-\left(x+1\right)y=0$ I am trying to solve $2x^2y''+3xy'-\left(x+1\right)y=0$ for a while but I can't make a progress.  
For a solution near $x=0$ I apply method of series expansion but it doesn't show any closed pattern to related $A_n$ to $A_0$ ; and by more 'direct' method (for $x \ne 0$) it can be reduced to $(x^{3/2} y')'=\dfrac{x+1}{2 \sqrt x} y$ but I don't know how to go further: On one hand there is $y$ is RHS and on the other hand, there is $x^{3/2}$ multiplied by $y'$ in LHS which both ways makes it impossible to proceed. 
However, there is a closed form (suggested by wolframalpha - without steps) for the solution : $$y(x) = c_1 \frac1x e^{\sqrt {2x}} (\sqrt 2 - 2 \sqrt x) + c_2 \frac1x e^{-\sqrt {2x}} ( 1+ \sqrt {2x}).$$
 [/INST] $x=0$ is a regular singular point.  The indicial equation is $r^2 + r/2 - 1/2 = 0$
which has roots $1/2$ and $-1$, so there are power series solutions of the form $\sum_{k=0}^\infty a_k x^{1/2+k}$ and $\sum_{k=0}^\infty b_k x^{-1+k}$.   The recurrences for $a_k$ and $b_k$ are
$$ \eqalign{a_{k+1} &= \frac{a_k}{(2k+5)(k+1)} \cr
             b_{k+1} &= \frac{b_k}{(2k-1)(k+1)} \cr}$$
Thus with $a_0 = 1$ the first series is
$$ \eqalign{&x^{1/2} \left(1 + \frac{x}{5 \cdot 1!}  + \frac{x^2}{7 \cdot 5 \cdot 2!} + \frac{x^3}{9 \cdot 7 \cdot 5 \cdot 3!} + \ldots\right)\cr&= x^{1/2} \left(1 + \frac{3 \cdot 2 x}{5 \cdot 3!} + \frac{3 (2 x)^2}{7 \cdot 5!} + \frac{3 (2x)^3}{9 \cdot 7!} + \ldots\right)\cr
&= \frac{3 \cosh(\sqrt{2x})}{2 x^{1/2}} - \frac{3 \sqrt{2} \sinh(\sqrt{2x})}{4 x}} $$
Similarly, with $b_0 = 1$ the second series turns out to be
$$ \frac{\cosh(\sqrt{2x})}{x} - \frac{\sqrt{2} \sinh(\sqrt{2x})}{\sqrt{x}}$$
</s>