Question

Find a complete and singular integrals of 2xz − px2 − 2qxy + pq = 0

Answer 1

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)