Find a complete and singular integrals of 2xz − px2 − 2qxy + pq = 0
Solution;
"f(x,y,z,p,q)=2xz-px^2-2qxy+pq=0.....(1)"
Now the charpits auxiliary equations are;
"\\frac{dp}{2z-2qy}=\\frac{dq}{0}=\\frac{dz}{px^2-pq+2xyq-pq}=\\frac{dx}{x^2-q}=\\frac{dy}{2x-p}"
The second fraction gives dq=0 ,which implies that q=a,a being an arbitrary constant.substitute in (1) we have;
"2zx-px^2-2axy+pa=0"
Which gives;
"p=\\frac{2x(z-ay)}{x^2-a}"
Substitute the values of p and q in "dz=pdx+qdy" we get;
"dz=\\frac{2x(z-ay)}{x^2-a}dx+ady"
Rewritten as;
"\\frac{dz-ady}{z-ay}=\\frac{2x}{x^2-a}dx"
After integration;
"log(z-ay)=log(x^2-a)+logb"
Or
"(z-ay)=b(x^2-a)"
From which;
"z=ay+b(x^2-a)"
(a and b are arbitrary constants)
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