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
Differentiate "(1)" partially with respect to "a"
"{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
Comments
The (x+f'(a)) was cancelled, but using ther property sqrt(t^2)=|t|=t*sign(t) for t=x+f'(a), the sign(t) cannot be cancelled because |t| and t can differ if t
I am really confused with the last line/final answer, shouldn't the (x+f'(a)) after sgn get cancelled? +- 2 sq rt
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