Answer to Question #199561 in Differential Equations for Rajkumar

Question

Find the integral surface of the PDE

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

Answer 1

QUESTION

Find the integral surface of the PDE 

$(𝑥 − 𝑦)𝑝 + (𝑦 − 𝑥 − 𝑧)𝑞 = z$

SOLUTION

By Lagrange’s method the auxiliary equations are as following:

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

One of a way to solve the system in symmetric form is to use the equal fractions property

$\frac{a_1}{b_1}=\frac{a_2}{b_2}=......=\frac{a_n}{b_n}=\frac{\lambda_1 a_1+\lambda_2 a_2+...+\lambda_n a_n}{\lambda_1 b_1+\lambda_2 b_2+...+\lambda_n b_n}$

Choosing $\lambda_1 =\lambda_2=\lambda_3=1$ as multipliers, each fraction on (1):

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

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

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

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

$x+y+z=c_1............................(2)$

Take the last two fractions of (1) and using (2) we get

$\frac{dy}{y-x-z}=\frac{dz}{z} ; x+y+z=c_1$

$\frac{dy}{y-(x+z)}=\frac{dz}{z} ; x+z=c_1-y$

$\frac{dy}{y-(c_1-y)}=\frac{dz}{z}$

$\frac{dy}{y-c_1+y}=\frac{dz}{z}$

$\frac{dy}{2y-c_1}=\frac{dz}{z}$

now integrate both side, we get,

$\frac{1}{2}.ln|2y-c_1|=ln|z|+ln|c_2|$

$ln|2y-c_1|=2.ln|z|+2.ln|c_2|$

$ln|2y-c_1|=2.ln|z|+ln|c_3|$

$ln|2y-c_1|-ln|z^2|=ln|c_3|$

$\frac{2y-c_1}{z^2}=c_3$

again put value of $c_1$

$\frac{2y-(x+y+z)}{z^2}=c_3$

$\frac{(y-x-z)}{z^2}=c_3.......................(3)$

Therefore, any integral surface of the differential equation

(𝑥 − 𝑦)𝑝 + (𝑦 − 𝑥 − 𝑧)𝑞 = 𝑧 is described by the equation

$f(c_1,c_3)=0\\ f(x+y+z,\frac{(y-x-z)}{z^2})=0\\$

where f is arbitrary function.