Answer to Question #105511 in Differential Equations for khushi

Question

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

Answer 1

$(x-y)p+(y-x-z)q=z\\ z=1, x^2+y^2=1$

Write the system in the symmetric form

$\frac{dx}{x-y}=\frac{dy}{y-x-z}=\frac{dz}{z}\\ i)\frac{dx}{x-y}=\frac{dy+dz}{y-x} \\ -dx=dy+dz\\ dx+dy+dz=0\\ d(x+y+z)=0\\ x+y+z=c_1\\ ii) x+z=c_1-y\\ \frac{dz}{z}=\frac{dy}{y-x-z}\\ \frac{dz}{z}=\frac{dy}{y-(x+z)}\\ \frac{dz}{z}=\frac{dy}{y-(c_1-y)}\\ \frac{dz}{z}=\frac{dy}{2y-c_1}\\ \int \frac{dz}{z}=\int\frac{dy}{2y-c_1}\\ \ln|z|+\frac{1}{2}\ln|c_2|=\frac{1}{2}\ln|2y-c_1|\\ c_2z^2=2y-c_1\\ c_2z^2=2y-x-y-z\\ \frac{y-x-z}{z^2}=c_2$

General solution is

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

Find particula solution

$\left\{\begin{matrix} z=1 \\ x^2+y^2=1\\ x+y+z=c_1\\ \frac{y-x-z}{z^2}=c_2 \end{matrix}\right.\\ \left\{\begin{matrix} x^2+y^2=1\\ x+y+1=c_1\\ y-x-1=c_2 \end{matrix}\right.\\ \left\{\begin{matrix} x^2+y^2=1\\ x^2+y^2+2xy=(c_1-1)^2\\ x^2+y^2-2xy=(c_2+1)^2 \end{matrix}\right.\\ \left\{\begin{matrix} 1+2xy=(c_1-1)^2\\ 1-2xy=(c_2+1)^2 \end{matrix}\right.\\ \left\{\begin{matrix} 2xy=(c_1-1)^2-1\\ 2xy=1-(c_2+1)^2 \end{matrix}\right.\\ (c_1-1)^2-1=1-(c_2+1)^2\\ (c_1-1)^2+(c_2+1)^2=2$

So particula solution is

$(x+y+z-1)^2+(\frac{y-x-z}{z^2}+1)^2=2$