Answer to Question #135034 in Differential Equations for Swathy

Question

Solve Z^3=pqxy

Answer 1

Using Charpit's method

we have the following equation,

$\frac{dp}{−pqy+2pz}=\frac{dq}{−pqx+2qz}=\frac{dz}{2pqxy}=\frac{dx}{qxy}=\frac{dy}{pxy}$

using multipliers p ,q, x & y in 1st, 2nd, 4th & 5th equations and equating it with equation 3rd.

$\frac{dz}{2pqxy}=\frac{(pdx+qdy+xdp+ydq)}{pqxy+pqxy−pqxy+2pxz−pqxy+2qyz}$

from equation

$z^3=pqxy$

$\frac{dz}{2z^3}=\frac{(pdx+qdy+xdp+ydq)}{2pxz+2qyz}$

$\frac{dz}{2z^3}=\frac{(d(px)+d(qy)}{2pxz+2qyz}$

$\frac{dz}{z^2}=\frac{d(px)+d(qy)}{px+qy}$

Integrate both the sides and we get

$-\frac{1}{z}=log(px+qy)+logc$

$\frac{1}{z}=-log((px+qy)c)$