Question

Find the complete integral of pq=1 by charpit's method

Answer 1

Let's solve this expression:

$f(x, y, z, p,q) = p + q - 3pq = 0$ 1)

Charpit's auxiliary equations for 1):

$\large\frac{dp}{\large\frac{\delta f}{\delta x} +p\large\frac{\delta f}{\delta z} }=\large\frac{dq}{\large\frac{\delta f}{\delta y} +q\large\frac{\delta f}{\delta z} }=\large\frac{dz}{-p\large\frac{\delta f}{\delta p} -q\large\frac{\delta f}{\delta q} }=\large\frac{dx}{\large\frac{-\delta f}{\delta p} }=\large\frac{dy}{\large\frac{\delta f}{\delta q}}$

$\large\frac{dp}{0+p.0} = \large\frac{dq}{0+q.0} + \large\frac{dz}{-p(1-3q)-q(1-3p)} = \large\frac{dx}{3q-1} = \large\frac{dy}{3p-1}$ 2)

Taking the first fraction of (2), we obtain $dp=0$

Integrating it, we get $p = a$

Substituting the value p = a in (1), we get $q = \large\frac{a}{3a-1}$

Now, putting the values of p and q respectively from(3) and (4) in $dz = pdx+qdy,$ we obtain

$dz= adx +$ $\frac{a}{3a-1} dy$

Integrating it, we obtain $z = ax + \large\frac{ay}{3a-1}+b$

Thus, the required comlete integral is $z = ax + \large\frac{ay}{3a-1}+b$