Answer to Question #199547 in Differential Equations for Rajkumar

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

general pde is given by Pp+Qq=R, we get

P=x-y

Q=y-x-z

R=z

By using lagrange method,

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

$\frac{dx+dy+dz}{x-y+y-x-z+z}=\frac{d(x+y+z)}{0}\newline x+y+z=c_1 .................2$

From 2, we get

$x+z=c_1-y\\ Then,\newline \frac{dz}{z}=\frac{dy}{y-x-z}\\ \frac{dz}{z}=\frac{dy}{y-(c_1-y)}\\ \frac{dz}{z}=\frac{dy}{2y-c_1}\\ \int \frac{dz}{z}=\int\frac{dy}{2y-c_1}\\ \ln|z|+\frac{1}{2}\ln|c_2|=\frac{1}{2}\ln|2y-c_1|\\ c_2z^2=2y-c_1\\ c_2z^2=2y-x-y-z\\ \frac{y-x-z}{z^2}=c_2...............3$

Therefore, the general solution is

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

Now, we find the particula solution,

From the given condition,

z=1and x²+y²=1.

Then,

equation 2 will be,

$x+y+1=c_1$ ...........4

and, equation 3 will be,

$\frac{y-x-1}{1^2}=c_2\newline y-x-1=c_2$ ..........5

Solving equation 4 and 5, we get

$x=\frac{c_1-c_2-2}{2} and y=\frac{c_1+c_2}{2}.............6\\ \text{put the value of x and y in} x^2+y^2=1.\\ (\frac{c_1-c_2-2}{2})^2+(\frac{c_1+c_2}{2})^2=1\\ (c_1-c_2-2)^2+(c_1+c_2)^2=4\\ (c_1-c_2)^2+4-4(c_1-c_2)+(c_1+c_2)^2=4\\ 2c_1^2+2c_2^2-4c_1+4c_2=0\\ c_1^2+c_2^2-2c_1+2c_2=0\\ \text{Again, put the value of} c_1 \text{and} c_2 \text{from equaton 2, 3.}\\ (x+y+z)^2+(\frac{x-y+z}{z^2})^2-2(x+y+z)+2(\frac{x-y+z}{z^2})=0\\ \newline (x+y+z)(x+y+z-2)+(\frac{x-y+z}{z^2})[\frac{x-y+z}{z^2}+2] =0$