Question

(2y^3-x^3)dx+3x^2ydy=0

Answer 1

$$(2y^3-x^3)dx+3x^2ydy=0$$

$$3\dfrac{y}{x}\cdot\dfrac{dy}{dx}+2\dfrac{y^3}{x^3}-1=0$$

Let $y=xv$

$$\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$$

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

$$3xv\dfrac{dv}{dx}=-2v^3-3v^2+1$$

$$\dfrac{3vdv}{-2v^3-3v^2+1}=\dfrac{dx}{x}$$

$$2v^3+3v^2-1=(v+1)^2(2v-1)$$

$$\dfrac{-3v}{2v^3+3v^2-1}=\dfrac{A}{2v-1}+\dfrac{B}{v+1}+\dfrac{C}{(v+1)^2}$$

$$A(v+1)^2+B(2v-1)(v+1)+C(2v-1)=-3v$$

$$v=\dfrac{1}{2}:\dfrac{9}{4}A=-\dfrac{3}{2}=>A=-\dfrac{2}{3}$$

$$v=-1: -3C=3=>C=-1$$

$$v=0:A-B-C=0=>B=A-C=\dfrac{1}{3}$$

$$-\dfrac{2}{3}\int\dfrac{dv}{2v-1}+\dfrac{1}{3}\int\dfrac{dv}{v+1}-\int\dfrac{dv}{(v+1)^2}$$

$$=\int\dfrac{dx}{x}$$

$$-\dfrac{1}{3}\ln|2v-1|+\dfrac{1}{3}\ln|v+1|+(\dfrac{1}{v+1})$$

$$=\ln|x|+\ln C$$

$$\dfrac{1}{3}\ln\dfrac{v+1}{2v-1}+\dfrac{1}{v+1}=\ln(Cx)$$

$$\dfrac{1}{3}\ln\dfrac{y+x}{2y-x}+\dfrac{x}{y+x}=\ln(Cx)$$