Determine the characteristics of the equation $z=p^2-q^2$, and find the integral surface which passes through the parabola $4z+x^2=0$, $y=0$.
I solved the question and my answer was:
The characteristics are
$x=2\lambda-\lambda e^{-t}$
$y=\sqrt{2}\lambda-\sqrt{2}\lambda e^{-t}$
$z=\frac{-\lambda^2}{4}e^{-2t}$
The Integral surface is $4z+(\sqrt{2}y-x)^2=0$
The answer in the textbook was:
The characteristics are
$x=2\lambda(2-e^{-t})$
$y=2\sqrt{2}\lambda(e^{-t}-1)$
$z={-\lambda^2}e^{-2t}$
The Integral surface is $4z+(\sqrt{2}y+x)^2=0$
$\lambda$ is a parameter. Hence the change in multiple of 2. However the integral surface was supposed to be the same.
I found the difference was due to the way i took $q_0$. After taking $x_0 = \lambda$, $y_0=0$, $z_0=-\frac{\lambda^2}{4}$
Solving for $p_0$ and $q_0$ accordingly,
$p_0=\frac{-\lambda}{2}$
$q_0^2=\frac{\lambda^2}{2}$
here when i take positive root of $q_0$, i get my solution. But when i take negative root, i get the one in textbook.
Are both solutions correct?. If not, why? My textbook does not show both solutions
Thanks in Advance