Answer to Question #154967 in Differential Equations for ANIK KUMAR GHOSH

Question

Solved by lagrange method

y^2 p^2+x^2q^2=x^2y^2z^2

Answer 1

$F(x,y,z,p,q)=y^2p^2+x^2q^2-x^2y^2z^2=0$

$\frac{dx}{F_p}=\frac{dy}{F_q}=\frac{dz}{pF_p+qF_q}=-\frac{dp}{F_x+pF_z}=-\frac{dq}{F_y+qF_z}$

$\frac{dx}{2py^2}=\frac{dy}{2qx^2}=\frac{dz}{2p^2y^2+2q^2x^2}=-\frac{dp}{2xq^2-2xy^2z^2-2pzx^2y^2}=-\frac{dq}{2yp^2-2yx^2z^2-2qzx^2y^2}$

$\frac{pdx+qdy-dz}{2p^2y^2+2q^2x^2-2p^2y^2-2q^2x^2}=\frac{pdx+qdy-dz}{0}=-\frac{dp}{2xq^2-2xy^2z^2-2pzx^2y^2}=-\frac{dq}{2yp^2-2yx^2z^2-2qzx^2y^2}$

$\frac{dx}{2py^2}=\frac{dy}{2qx^2}=\frac{pdx+qdy-dz}{0}$

$p=0, q\neq0$

Substitute this result into the initial equation, and we get:

Then:

$x^2q^2=x^2y^2z^2$

$q=yz$

$\frac{y^2}{2}=lnz+c_1$

$p\neq0, q=0$

Then:

$p=xz$

$\frac{x^2}{2}=lnz+c_2$

The general solution:

$F(c_1,c_2)=F(\frac{y^2}{2}-lnz, \frac{x^2}{2}-lnz)$