Solve the differential equation: $(3x^2-y^2)dy-2xdx=0$

Question

I have the differential equation: $$(3x^2-y^2) dy - 2x dx=0 $$

I need it to look like the equation: $$y'+p(x)y=q(x)$$ in order to apply the integrating factor.

I believe it can be solved using integrating factors because that is the section in the textbook it has come from, but I am totally stumped on how to get $y'$ isolated for the integrating factor.

Is this an integrating factor equation, and if so how do I deal with these types of equations?

Answer 1

I have the differential equation: $$(3x^2-y^2) \ dy - 2x \ dx=0 $$

Multiply the integral factor: $e^{-3y}$, then we get:

$$-2xe^{-3y}dx+(3x^2-y^2)e^{-3y}dy=0$$

Now this equation becomes exact.

Answer 2

With $x^2=z$ you could write this as $\dfrac{\mathrm{dz}}{\mathrm{dy}}=3z-y^2$ which is in linear form