Question

Find the general integral of the equation

(x-y)p+(y-x-z)q=z,

And the particular solution through the circle z=1,x²+y²=1

Answer 1

Comparing with $Pp+Qq=R,$ we get

$P=x-y, Q=y-x-z, R=z$

The corresponding system of equations is

$$\dfrac{dx}{x-y}=\dfrac{dy}{y-x-z}=\dfrac{dz}{z}$$

$$\dfrac{dx+dy+dz}{x-y+y-x-z+z}$$

$$\dfrac{dx+dy+dz}{0}=\dfrac{d(x+y+z)}{0}$$

$$d(x+y+z)=0$$

$x+y+z=c_1$ is the first independent integral.

Consider

$$\dfrac{dx-dy+dz}{x-y-y+x+z+z}$$

$$\dfrac{dx-dy+dz}{2(x-y+z)}=\dfrac{d(x-y+z)}{2(x-y+z)}$$

$$\dfrac{d(x-y+z)}{2(x-y+z)}=\dfrac{dz}{z}$$

$$\dfrac{d(x-y+z)}{x-y+z}=2\cdot\dfrac{dz}{z}$$

$$\ln(x-y+z)=\ln(z^2)+\ln c_2$$

$\dfrac{x-y+z}{z^2}=c_2$ is the second independent integral.

The general solution is

$$\phi(x+y+z, \dfrac{x-y+z}{z^2} )=0$$

Let $x+y+z=t$ and $\dfrac{x-y+z}{z^2} =u.$

$$\phi(t, u)=0$$

For $z=1$ we get $t=x+y+1$ and $u=\dfrac{x-y+1}{1^2}$

Then

$$x=\dfrac{t+u}{2}-1, y=\dfrac{t-u}{2}$$

Given $x^2+y^2=1$

$$(\dfrac{t+u}{2})^2-(t+u)+1+(\dfrac{t-u}{2})^2=1$$

$$t(t-2)+u(u-2)=0$$

Hence

$$(x+y+z)(x+y+z-2)$$

$$+(\dfrac{x-y+z}{z^2})(\dfrac{x-y+z}{z^2}-2)=0$$