Then the equation is $g^2=\frac{dg}{dx}$ which has solution $g(x)=\frac{-1}{x+c}$. Integrate that to get the solution $f(x)=-\log(|x+c_1|)+c_2$. $\endgroup$ –

Exact Differential Equations. An equation of the form: M(x, y) dx + N(x, y) dy = 0 is exact if: \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} The solution is obtained by the integrating the M with the respect to the x and N with the respect to y. Read More about Exact Differential Equations. Numerical Methods

When n = 0 the equation can be solved as a First Order Linear Differential Equation. When n = 1 the equation can be solved using Separation of Variables. For other values of n we can solve it by substituting u = y 1−n and turning it into a linear differential equation (and then solve that).

To solve ordinary differential equations (ODEs), use methods such as separation of variables, linear equations, exact equations, homogeneous equations, or numerical methods.

$Ay'' + B(y')^2 + C = 0$ where $A, B, C$ are constants. I checked that 2nd order equations of the form $Ay'' + By' + Cy$ can be treated as $Ay^2 + By + C = 0$ and then proceeded further. But how a...