Answer to Question #237499 in Differential Equations for James

Question

Solve the differential equation of the following homogeneous equation.

(x^2 - xy + y^2)dx - xydy = 0

Answer 1

$(x^2 -xy +y^2)dx -xydy =0 \\ Divide \ by \ x^2 \\ (1-\frac{y}{x} +\frac{y^2}{x^2}) dx - \frac{y}{x} dy = 0 \\ Let \ \frac{y}{x} = v \\ y=vx \\ (1-v+v^2) dx - v dy = 0 \\ dy = (vdx+xdv) \\ (1-v+v^2) dx - v (v dx+x dv) = 0 \\ (1-v+v^2-v^2) dx - vx dv = 0 \\ (1-v) dx - vx dv = 0 \\ (1-v) dx = vx dv \\ Divide \ both \ sides \ by \ (1-v) x\\ \frac{dx}{x} = \frac{v}{(1-v)} dv \\ Integrate \ both \ sides\ , we \ get: \\ \frac{v}{(1-v)} = \frac{1}{(1-v)} - 1 \\ ∫ \frac{dx}{x} = ∫ \frac{dv}{(1-v)} - ∫ dv \\ Substitute \ v=\frac{y}{x}\\ ln(x) + C_1 = -ln(1-v) - v \\ ln(x) + C_1 = - ln[\frac{(x-y)}{x}] - \frac{y}{x}\\ ln[\frac{(x-y)}{x}] + \frac{y}{x} = - ln(x) + C$