Given that z=ax+a2−4y+c is the complete integral of the PDE, p2−q2=4
To find the general solution we put c=f(a), where f is an arbitrary function
z=ax+a2−4y+f(a) (1)Differentiate (1) partially with respect to a
x+2a2−4y2a+f′(a)=0a2−4ya=−(x+f′(a))a2=a2(x+f′(a))2−4y(x+f′(a))2a=±2(x+f′(a))2−1(x+f′(a))2yz=ax−x+f′(a)a+f(a)The general solution is
z=±2(x+f′(a))2−1(x+f′(a))2yx∓sgn(x+f′(a))2(x+f′(a))2−1y+f(a)
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