Answer to Question #283471 in Differential Equations for Ruchi

Question

Shiw that the equations xp-yp=0, z(xp+yq)=2xy are compatible and solve them

Answer 1

Solution:

Here, we may take $f=x p-y q, g=z(x p+y q)-2 x y$ so that

$\dfrac{\partial(f, g)}{\partial(x, p)}=2 x y, \quad \dfrac{\partial(f, g)}{\partial(z, p)}=-x^{2} p, \quad \dfrac{\partial(f, g)}{\partial(y, q)}=-2 x y, \quad \dfrac{\partial(f, g)}{\partial(z, q)}=x y p$

from which it follows that

$[f, g]=x p(y q-x p)=0$ (here [f,g] is the J, i.e. jacobian)

since $x p=y p$ .

The equations are therefore compatible.

It is readily shown that $p=y / z, q=x / z$ , so that we have to solve

which has solution

$z d z=y d x+x d y$

where $c_{1}$ is a constant.

$z^{2}=c_{1}+2 x y$