# 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.?

##### 1 Answer
Aug 11, 2018

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)}