Question

Find the general integral of the equation (x-y)p+(y-x-z)q=z and particular solution through the circle z=1,x^2+y^2=1.?

Answer 1

Making the change of variables

# r = \sqrt{x^2+y^2} #
# \theta = \arctan(\frac{y}{x}) #

we obtain

# (1-\frac{z}{r}\cos\theta-\sin(2\theta))u_r +(r\cos(2\theta)-2\sin\theta-z)u_{\theta} = 0 #

now considering `z = 1` the PDE can be solved using the characteristic method. as

# \frac{dr}{(1-\frac{z}{r}\cos\theta-\sin(2\theta))} = \frac{d\theta}{(r\cos(2\theta)-2\sin\theta-z)} #