Hey student,

Since this is a homogeneous differential equation, let y = xvy = xv . As dy = vdx + xdvdy = vdx + xdv , we obtain

(x−vx)dx + (x+vx)(vdx+xdv) = 0

(x−vx)dx + (x+vx)(vdx+xdv) = 0.

This simplifies to

(1+v2)dx + x(1+v)dv = 0

(1+v2)dx + x(1+v)dv=0

Dividing both sides by x(1+v2)x(1+v2) :

dxx+1+v1+v2dv=0

dxx+1+v1+v2dv=0

Now, we can integrate both sides:

ln|x| + arctanv + 12ln(1+v2) = A

ln|x| + arctanv + 12ln(1+v2) = A.

To make this solution look a little nicer, multiply both sides by 2 (and let C=2AC=2A ):

2ln|x|+ 2arctanv + ln(1+v2) = C

2ln|x|+ 2arctanv + ln(1+v2) = C.

Rewrite this in terms of yy :

ln(x2) + 2arctan(yx) + ln(1+y2x2) = C

ln(x2) + 2arctan(yx) + ln(1+y2x2) = C

Finally, we use the product rule for logarithms:

2arctan(yx) + ln(x2+y2) = C

I hope you find it useful.