Homogeneous Differencial Equation

Question

[x-yarctan(y/x)]dx + [xarctan(y/x)]dy= 0

Answer 1

this is how I approach

first divide by x*dx x not zero

set v=y/x so dy/dx=v+xdv/dx

have v*tan^-1v-1=tan^-1v*(v+x*dv/dx)

which is separable.

need to integrate tan^-1v*dv on one side and (-1/x)dx on other side

actually, tan^-1v=1-(v³)/3+(v⁵)/5-/+...

for v between and including -1 to 1

so solution would be y/x-(y/x)⁴/4*3+(y/x)⁶/6*5-/+...=-ln(x)+C

for y/x between and including -1 and +1

Answer 2

dy/dx = [yarctan(y/x) - x]/[xarctan(y/x)]

y = xv; dy/dx = v + xv';

v + xv' = [xvarctanv - x]/[xarctanv;

v + xv' = v - 1/arctanv; xdv/dx = - 1/arctanv; artanvdv = - dx/x; ∫arctanv = - ∫dx/x; varctanv - 1/2ln(1 + v²) =

-lnIxI + C. But v = y/x. So, y/xarctan(y/x) - 1/2ln(1 + y²/x²) = - lnIxI + C