Use Abel's Theorem: if your equation is of the form $y''+p(t)y'+q(t)y=g(t)$, then the Wronskian of your solution set is given by $W=e^{\int -p(t)dt}$. We also know the Wronskian of two solutions is given by $y_1y_2'-y_2y_1'$, so setting those two equal gives a simple first order equation. In this case, rewriting your equation in standard form gives us
$$y''+0y'+(0.1875x^{-2}-x^{-1})y=0$$
(note ever symbol is a + and the leading coefficient of $y''$ is a $1$, that's standard form)
so we get $W=e^{\int 0dx}=e^C$ We might as well pick the simplest $C$ and go for $C=0$, so we get $W=1$.
Then, we get, choosing $y_1=x^{1/4}e^{2\sqrt x},y_1'=\frac 1 4x^{-\frac 3 4}e^{2\sqrt x}+x^{1/4}e^{2\sqrt x}x^{-1/2}=e^{2\sqrt x}(\frac 1 4x^{-\frac 3 4}+x^{-\frac 1 4})$
So, $1=x^{1/4}e^{2\sqrt x}y_2'-e^{2\sqrt x}(\frac 1 4x^{-\frac 3 4}+x^{-\frac 1 4})y_2$. Divide through by the coefficient of $y_2'$ to put this into standard first order form, and change the - to a +, to get
$y_2'+(-\frac 1 4x^{-1}-x^{-\frac 1 2})y_2=x^{-\frac 1 4}e^{-2\sqrt x}.$
So our integrating factor is $\mu=e^{\int-\frac 1 4x^{-1}-x^{-\frac 1 2}dx}=e^{-\frac 1 4\ln x-2x^\frac 1 2}=x^{-\frac 1 4}e^{-2x^{\frac 1 2}}$
You'll note that the integrating factor is exactly $(y_1^{-1})$. This is not a coincidence, it should always work that way. If you don't get that, you made a mistake setting up your problem.
So, setting up your answer, we have $y_2=\frac {\int g(t)\mu (t)dt} {\mu (t)}$, so $y_2=\frac {\int x^{-\frac 1 4}e^{-2x^{\frac 1 2}}x^{-\frac 1 4}e^{-2\sqrt x}dx}{x^{-\frac 1 4}e^{-2x^{\frac 1 2}}}=\frac {\int x^{-\frac 1 2}e^{-4\sqrt x}dx}{x^{-\frac 1 4}e^{-2x^{\frac 1 2}}}$
Now, setting $u=-4\sqrt x$, $du=-2x^{-\frac 1 2}dx,-\frac 1 2 du=x^{-\frac 1 2}dx$we have the integral in the top converts to $\int -\frac 1 2 e^u du=-\frac 1 2 e^u +C=-\frac 1 2 e^{-4\sqrt x} +C$
Plugging that back into the top, we get $y_2=\frac {-\frac 1 2 e^{-4\sqrt x} +C}{x^{-\frac 1 4}e^{-2x^{\frac 1 2}}}=-\frac 1 2e^{-2\sqrt x}x^{-\frac 1 4}+Cx^{\frac 1 4}e^{2\sqrt x}$
Now, the second part with the C is just your original $y_1$ solution times an arbitrary constant. The $-\frac 1 2$ came up in this calculation because we CHOSE an arbitrary $C$ when we calculated the wronskian, any other constant multiple will also work, so we might as well take the constant multiple to be 1, and that gets to the second solution you have.
Best Answer
There is no general approach for all ODE's of the form $ y'' = f(y) $. In fact, even when the functions $f(y)$ are restricted to relatively simple functions, the solutions can be very exotic. For example, if $f(y) = 2k^3 y^3-(1+k^2)y$ where $k$ is some fixed constant, the resulting solution (with initial conditions $y(0)=y'(0)=0$ ) is something called the Jacobi elliptic function $\operatorname{sn}(z,k)$. That's just one case for what is a relatively simple $f(y)$ (it was a 3rd degree polynomial). People devote a lot of effort into studying methods to solve DE's like yours, for specific special families of functions $f(y)$, with each special family having special techniques to solve them.