Solve the differential equation subject to the indicated con

Solution

let y = vx ... this is a technique whenever the terms all have the same degree... then dy = v dx + x dv § thus v^2 x^2 dx + (x^2 + vx^2 + v^2x^2) [v dx + x dv] = 0 the above equation is now separable x^2 [v + 2v^2 + v^3] dx + x^3 [1 + v + v^2] dv = 0 {-1/x} dx = [1 + v + v^2]/[v + 2v^2 + v^3] dv ... by partial fractions ... {-1/x} dx = [1/v - 1/(1+v)^2] dv -ln x = ln v + 1/(1+v) + C v = y/x ln|x| + ln|y/x| + 1/(1 + y/x) + C ln|y| + x/(x+y) + C = 0 ... C = 0 if x = 0 & y = 1 (x+y) ln|y| + x = 0 ... whenever cross multiplied.. the rightside is still 0.