Answer to Question #299382 in Differential Equations for Pankaj

Question

Find the integral surface of the PDE:

x²p + y²q +z² =0

which passes through the hyperbola

xy =x+y, z=1.

Answer 1

The auxiliary equation is $\dfrac{dx}{x^2}=\dfrac{dy}{y^2}=\dfrac{dz}{-z^2}.$

Taking the first two ratios,

$\begin{aligned} \dfrac{dx}{x^2}&=\dfrac{dy}{y^2}\\ \text{Integrating both sides, we get}\\ \dfrac{-1}{x} &= \dfrac{-1}{y} + c_{1}\\ \dfrac{1}{y} -\dfrac{1}{x} &= c_{1}\qquad\qquad(1)\\ \end{aligned}$

Taking last two ratios,

$\begin{aligned} \dfrac{dy}{y^2}&=-\dfrac{dz}{z^2}\\ \text{Integrating both sides, we get}\\ \dfrac{-1}{y} &= \dfrac{1}{z} + c_{2}\\ -\dfrac{1}{y} -\dfrac{1}{z} &= c_{2}\qquad\qquad(2)\\ \end{aligned}$

The integral curve is $\phi(c_1,c_2) = 0$, i.e., $\phi(\frac{1}{y}-\frac{1}{x},-\frac{1}{y}-\frac{1}{z}) = 0$.

Given that the integral surface passes through $xy=x+y, z=1$.

Letting $x(s) = s$ as the parametric representation, we get

$\begin{aligned} sy &= s+y\\ sy-y&=s\\ y(s-1) &= s\\ \therefore y(s) &= \dfrac{s}{s-1} \end{aligned}$

Using $x(s) = s, y(s) = \dfrac{s}{s-1}, z=1$ in (1) and (2) we get

$\begin{aligned} \frac{s-1}{s}-\frac{1}{s} &= c_1 \Rightarrow c_1 = \dfrac{s-2}{s}\Rightarrow c_1 = 1- \dfrac{2}{s} \Rightarrow \dfrac{1}{s} = \dfrac{1-c_1}{2}\qquad (3)\\ -\left(\frac{s-1}{s}\right)-1 &= c_2\Rightarrow c_2 = \dfrac{1-2s}{s} \Rightarrow c_2 = \dfrac{1}{s}-2 \qquad\qquad\qquad\qquad(4)\\ \end{aligned}$

Using (3) in (4),

$\begin{aligned} c_2 &= \dfrac{1-c_1}{2} - 2\\ 2c_2 - 1 + c_1+4 &= 0\\ c_1 + 2c_2+3 &=0\\ \left(\dfrac{1}{y}-\dfrac{1}{x}\right) - 2\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+3&=0~~~(\text{Using (1) \& (2)})\\ \dfrac{x-y}{xy}-2\left(\dfrac{z+y}{yz}\right)+3&=0\\ xz-yz-2xz-2xy+3xyz &=0\\ 3xyz-2xy-xz-yz&=0 \end{aligned}$

which is the required integral surface.