Solve a special differential equation How can I rigorously solve this differential equation ? 
$$\frac{dy}{dx}=a y^{-1/6}$$ 
I know how to solve linear ODEs but I don't know how to do it with this one.
 A: Anytime you want to solve an equation of the type $$y'=f(x)g(y),$$
you follow the following steps:
$$\frac{dy}{dx} = f(x)g(y)\\
\frac{dy}{g(y)} = f(x) dx\\
\int\frac{dy}{g(y)} = \int f(x) dx$$
While the second step in this derivation is fishy (and, to some extend even meaningless), it is a simple way to remember how to get the last row, which is again correct (its correctnes can be proven through rigorous means). Calculating the integral will result in an equation $$G(y) = F(x) + C.$$ Usually, this equation can be solved for $y$.
A: If $a$ is constant, then follow this procedure
\begin{align}
\frac{dy}{dx}&=a y^{-1/6} \\
\frac{dy}{y^{-1/6}}&=a \, dx & \text{separate variables}\\
y^{1/6} \, dy &= a\, dx \\
\int y^{1/6} \, dy &= \int a\, dx \\
\frac{6}{7} y^{7/6} &= ax+C \\
\end{align}
Now try to get $y$ by itself, so you can get your solution explicity.
However, if $a$ depends on $x$, then $a=a(x)$, unless we know a particular function based on $x$. We would need to replace all the $a$'s with $a(x)$'s:
\begin{align}
\frac{dy}{dx}&=a(x) y^{-1/6} \\
\int y^{1/6} \, dy &= \int a(x)\, dx \\
\frac{6}{7} y^{7/6} &= \int a(x)\, dx+C \\
\end{align}
Once again, try to get $y$ by itself. If you have a particular $a(x)$ function that you can integrate (e.g. $a(x)=x^2$), then evaluate $\int a(x) \, dx$ accordingly. Otherwise, leave it as that expression.
