The general (nonzero) solution of the homogeneous differential?

Question

The general (nonzero) solution of the homogeneous differential
equation
(x2+y2)dx+2xydy = 0
can be written as f (x;y)+x3 =C
f (x;y) = (Note: the notation for absolute value of a number y is abs(y)

Answer 1

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

dy/dx = -(x^2+y^2)/(2xy)

dy/dx = (-1/2){1+(y/x)^2}/(y/x) ....... (1)

Let y/x = v so y = vx so dy/dx = v + x(dv/dx).

Substituting in (1)

v + x(dv/dx) = (-1/2)(1+v^2)/v

x(dv/dx) + (1+3v^2)/(2v) = 0

[2v/(1+3v^2)]dv + (1/x)dx = 0

Integrating
∫[2v/(1+3v^2)]dv + ∫(1/x)dx = 0

(1/3)∫[6v/(1+3v^2)]dv + ∫(1/x)dx = 0

(1/3)∫[1/(1+3v^2)]d(1+3v^2) + ∫(1/x)dx = 0

(1/3)ln(1+3v^2) + lnx = lnk where k=arbitrary const. of integration

ln(1+3v^2) + ln(x^3) = ln(k^3)

ln{(1+3v^2)(x^3)/(k^3)} = 0

(1+3v^2)(x^3)/(k^3) = 1

{1+3(y/x)^2}(x^3) = k^3

(x^3 + 3xy^2) = k^3

Let k^3 = c

Then reqd. solution is x^3 + 3xy^2 = c.

Answer 2

psb is right, except x should be replaced with abs(x).

minor mistake.

give him the ten points.

Steve