Answer to Question #115582 in Differential Equations for Neha

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

comparing with Pp+Qq=R, we get

P=x-y

Q=y-x-z

R=z

the corresponding system of equations is

$\frac{dx}{x-y}=\frac{dy}{y-x-z}=\frac{dz}{z}$ ......1

adding all the above equations in 1

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

d(x+y+z)=0

x+y+z=f is the first independent integral

now considering 1 again,

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

$d[\frac{(ln (x-y+z))}{z^2}]=0$

g= (x-y+z)/z² is the second independent integral

the general solution is

U(P,Q)=0 WHERE P= x+y+z and Q=x-y+z/z² ....2

since we have z=1

P= x+y+1 and Q=x-y+1

so, on solving

x= $\frac{P+Q}{2}-1 and y= \frac{P-Q}{2}$

Now, given x²=y²=1

putting values of P and Q,

P(P-2)+Q(Q-2)=0

U(P,Q)= P(P-2)+Q(Q-2)

Putting the values of P and Q from 2

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