Question

(2xy+3y2)dx-(2xy+X2)dy=0

Answer 1

$$- \frac{{dy}}{{dx}}( {x^2} + 2xy) +2xy+ 3{y^2} = 0$$

Let $y(x) = xv(x)$, which gives $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$ :

$$-(v+x\dfrac{dv}{dx})( x^2 + 2x^2v) +2x^2v+3x^2v^2 = 0$$

Simplify:

$$x^2(-x\dfrac{dv}{dx}(1+2v)-v-2v^2+2v+3v^2)=0$$

$$\dfrac{1+2v}{v(v+1)}dv=\dfrac{dx}{x}$$

$$(\dfrac{1}{v}+\dfrac{1}{v+1})dv=\dfrac{dx}{x}$$

Integrate

$$\int(\dfrac{1}{v}+\dfrac{1}{v+1})dv=\int\dfrac{dx}{x}$$

$$\ln(|v|)+\ln(|v+1|)=\ln(|x|)+\ln c_1$$

$$v(v+1)=c_1x$$

$$xv(xv+x)=c_1x^3$$

$$y^2+xy=c_1x^3$$