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Find the integral surface of the differential equation $(x-y)p+(y-x-z)q=z$ passing through the circle C: $z=1, x^2+y^2=1$

Clearly the Lagrange's auxillary equations are

$\frac{dx}{P} = \frac{dy}{Q}. =\frac{dz}{R}$

Where P=$(x-y)$ ,Q=$y-x-z$ & R=$z$

on comparing the given P.D.E with the general quasilinear equation P(x,y,z) p+Q(x,y,z)q=R(x,y,z)

I obtained two solutions

$x+y+z=c_1$ & $\frac{x-y+z}{z^2}=c_2$

Then I substituted $x=s$ where S is a parameter. So $y=\sqrt{1-s^2}$ & $z=1$. I need to find a relation in $c_1$ &$c_2$ Then substitute for $c_1$ &$c_2$ to find the final integral surface passing through given circle. How can I proceed now?

Kavita
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1 Answers1

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Your calculus is correct. The characteristic equations are : $$\begin{cases} x+y+z=c_1\\ \frac{x-y+z}{z^2}=c_2 \end{cases}$$ The general solution of the PDE, expressed on the form of implicit equation, is : $$F(X,Y)=0 \quad \begin{cases} X=x+y+z\\ Y=\frac{x-y+z}{z^2} \end{cases}$$ where $F(X,Y)$ is any differentiable equation of two variables.This function has to be determined according to the conditions :

First condition : $z=1 \begin{cases} X=x+y+1\\ Y=x-y+1 \end{cases} \quad\to\quad \begin{cases} x=\frac{X+Y}{2}-1\\ y=\frac{X-Y}{2} \end{cases}$

Second condition : $x^2+y^2=1 \quad\to\quad \left(\frac{X+Y}{2}-1\right)^2+\left(\frac{X-Y}{2}\right)^2=1$

After simplification : $X(X-2)+Y(Y-2)=0$ which determines the function $$F(X,Y)=X(X-2)+Y(Y-2)$$ Thus, with $\begin{cases} X=x+y+z\\ Y=\frac{x-y+z}{z^2} \end{cases}$ the particular solution according to the specified condition is : $$(x+y+z)(x+y+z-2)+\frac{x-y+z}{z^2}\left(\frac{x-y+z}{z^2}-2\right)=0$$ This is the equation of the surface passing through the specified circle.

JJacquelin
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