Answer to Question #105488 in Differential Equations for khushi

Question #105488
Given that z = ax + √a² − 4y + c 2 is the complete integral of the PDE, p² − q² =4 determine its general integral.
1
Expert's answer
2020-03-19T05:28:55-0400

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)"


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Comments

Assignment Expert
25.04.20, 00:27

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

Erina
24.04.20, 21:05

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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