Asymptotic solution to a differential equation near zero

Question

I am trying to get the both the asymptotic solutions of the equation $y''(x)=\sqrt{x} \cdot y(x)$ as $x\rightarrow 0$. But when I put $y(x)=\exp(S(x))$ since $x=0$ is an irregular singular point, no matter whether I neglect $S''$ or $S'^{2}$, the approximation does not seem to be valid after evaluating the solutions unless I ignore the constant of integration. How can I proceed to get asymptotic expansion of both the solutions? Please provide hints.

Answer 1

The general solution, according to Maple, is

$$ y(x) = c_1 \sqrt{x}\; I_{2/5}\left(\dfrac{4}{5} x^{5/4}\right) + c_2 \sqrt{x} \;K_{2/5}\left(\dfrac{4}{5} x^{5/4}\right)$$ where $I_{2/5}$ and $K_{2/5}$ are modified Bessel functions of the first and second kinds.

The series expansions at $x=0$ (for $x > 0$), again according to Maple, are

$$ \eqalign{ \sqrt{x}\; I_{2/5}\left(\dfrac{4}{5} x^{5/4}\right) &= {\frac {{2}^{{ {9}/{10}}}{5}^{3/5}\Gamma \left( 3/5 \right) \sqrt {5+\sqrt {5}}}{8 \pi }} x+{\frac {{2}^{{ {9}/{10}}}{5}^{3/5} \Gamma \left( 3/5 \right) \sqrt {5+\sqrt {5}}}{70\,\pi }} x^{7/2}\cr &+{ \frac {{2}^{{ {9}/{10}}}{5}^{3/5}\Gamma \left( 3/5 \right) \sqrt {5+\sqrt {5}}}{2100\,\pi }} x^6+O \left( {x}^{17/2} \right) \cr \sqrt{x} \;K_{2/5}\left(\dfrac{4}{5} x^{5/4}\right) &= -{\frac {\pi \,{2}^{1/10}{5}^{2/5}\sqrt {5+\sqrt {5}} \left( -5+ \sqrt {5} \right) }{20 \;\Gamma \left( 3/5 \right) }}-\dfrac{{2}^{2/5}{5}^{3 /5}\Gamma \left( 3/5 \right)}{4} x\cr &-{\frac {\pi \,{2}^{1/10}{5}^{2/5}\sqrt {5+\sqrt {5}} \left( -5+\sqrt {5} \right) }{75\,\Gamma \left( 3/5 \right) }} x^{5/2}-\dfrac{{2}^{2/5}{5}^{3/5}\Gamma \left( 3/5 \right)}{35} {x}^{7/2}\cr &-{\frac {\pi \,{2}^{1/10}{5}^{2/5}\sqrt {5+ \sqrt {5}} \left( -5+\sqrt {5} \right) }{1500\,\Gamma \left( 3/5 \right) }} x^5-{\frac {{2}^{2/5}{5}^{3/5}\Gamma \left( 3/5 \right) }{1050}} x^6\cr &-{\frac {\pi \,{2}^{1/10}{5}^{2/5}\sqrt {5+\sqrt { 5}} \left( -5+\sqrt {5} \right) }{73125\,\Gamma \left( 3/5 \right) }} x^{15/2} +O \left( {x}^{17/2} \right) }$$

So Antonio's series are correct. You might note, BTW, that the change of dependent variable $ x = t^2$ gives you a differential equation $$ t y'' - y' - 4 t^4 y $$ which has a regular singular point at $t=0$, with indicial equation $r^2 - 2 r = 0$, and the recursion for coefficients $$ a_{i+5} = \dfrac{4}{(i+5)(i+3)} a_i$$ where $$ y = a_0 + a_5 t^5 + a_{10} t^{10} + \ldots = a_0 + a_5 x^{5/2} + a_{10} x^5 + \ldots$$ or $$ y = a_2 t^2 + a_7 t^7 + a_{12} t^{12} + \ldots = a_2 x + a_7 x^{7/2} + a_{12} x^6 + \ldots$$