Question

Show that the equation xp =ya and z(xp +yq) = 2xy are compatible and solve them​

Answer 1

Answer:

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Answer 2

Answer:

The equation $xp =ya$and $z(xp +yq) = 2xy$ are compatible

Step-by-step explanation:

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

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

from which it follows that

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

since $x p=y p$.

The equations are therefore compatible.

It is readily shown that $p=\frac{y}{z}, q=\frac{x}{z}$, so that we have to solve

which has solution

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

where $c_{1}$ is a constant.

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