Question

Given that z = ax +sqrt (a^2 − 4y) + c is the complete integral of the PDE, p^2 − q^2 =4 determine its general integral.

Answer 1

Given that $z=ax+\sqrt{a^2-4y}+c$ is the complete integral of the PDE, $p^2-q^2=4$

To find the general solution we put $c=f(a),$ where $f$ is an arbitrary function

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ z=ax+\sqrt{a^2-4y}+f(a) \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

Differentiate $(1)$ partially with respect to $a$

$$x+{2a \over 2\sqrt{a^2-4y}}+f'(a)=0$$

$${a \over \sqrt{a^2-4y}}=-(x+f'(a))$$

$$a^2=a^2(x+f'(a))^2-4y(x+f'(a))^2$$

$$a=\pm2\sqrt{{(x+f'(a))^2y \over (x+f'(a))^2-1}}$$

$$z=ax-{a \over x+f'(a)}+f(a)$$

The general solution is

$$z=\pm2\sqrt{{(x+f'(a))^2y \over (x+f'(a))^2-1}}x\mp\text{sgn}(x+f'(a))2\sqrt{{y \over (x+f'(a))^2-1}}+f(a)$$