Answer to Question #269971 in Differential Equations for rwong

Question

Differential Equation: y^2dx-x(2x+3y)dy=0

Answer 1

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

$$y'=\dfrac{y^2}{x(2x+3y)}$$

Substitute $y=vx$

$$y'=v+v'x$$

$$v+v'x=\dfrac{v^2x^2}{x(2x+3vx)}$$

$$v'x=\dfrac{v^2-2v-3v^2}{(2+3v)}$$

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

Integrate

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

$$\dfrac{2+3v}{v(v+1)}=\dfrac{A}{v}+\dfrac{B}{v+1}=\dfrac{Av+A+Bv}{v(v+1)}$$

$$A=2, B=1$$

$$\int\dfrac{2}{v}dv+\int\dfrac{1}{v+1}dv=-\int\dfrac{2dx}{x}$$

$$2\ln(|v|)+\ln(|v+1|)=-2\ln(|x|)+\ln C$$

$$v^2(v+1)=\dfrac{C}{x^2}$$

$$x^2v^2(vx+x)=Cx$$

$$y^2(x+y)=Cx$$