Question

Find the integral surface for the differential equation z(xz_x-yz_y)=y²-x² passing through the initial data curve (2s, s, s)

Answer 1

$z(xz_x-yz_y)=y^2-x^2\\ xz_x-yz_y=\frac{y^2-x^2}{z}\\ P=x , Q=-y, R=\frac{y^2-x^2}{z}\\ \frac{dx}{dt}=P=x\\ \frac{dx}{x}=dt \text{ Integrate}\\ \ln x=t+c_1\\ x=C_1e^t\\ \text{From the initial curve, we know that } x(0)=2s\\ C_1=2s\\ x=2se^t....................(1)\\ \text{Similarly, }\\ \frac{dy}{dt}=Q=-y\\ -\ln y=t+c_2\\ y=C_2e^{-t}\\ y(0)=s\\ C_2=s\\ y=se^{-t}................(2)\\ \text{Also,}\\ \frac{dz}{dt}=R=\frac{y^2-x^2}{z}\\ zdz=(y^2-x^2)dt\\ \frac{z^2}{2}=(y^2-x^2)t+c_3\\ z(0)=s\\ c_3=\frac{s^2}{2}\\ z^2=2(y^2-x^2)t+s^2...............(3)$

We need to get the value for $s \text{ and }t$

From (1) and (2),

$\frac{x}{y}=2e^{2t}\\ t=\ln(\frac{x}{2y})^2$

Substitute this into (1)

$x=2se^{\ln (\frac{x}{2y})^2}\\ x=\frac{2s.x^2}{4y^2}\\ s=\frac{2y^2}{x}$

Put the values if s and t into (3) to get the solution.

$z^2=2(y^2-x^2)\ln (\frac{x}{2y})^2+\frac{4y^4}{x^2}\\ x^2z^2=2x^2(y^2-x^2)\ln (\frac{x}{2y})^2+4y^4$

Thus is the solution and the characteristics are

$x=\frac{2y^2}{s}$