Solve yz dx - z^2 dy - xy dz = 0

Solve \[ yz\,dx - z^2\,dy - xy\,dz = 0. \]

Step 1 · Check exactness

With $P=yz$, $Q=-z^2$, $R=-xy$: \[ P_y=z,\ Q_x=0;\qquad P_z=y,\ R_x=-y;\qquad Q_z=-2z,\ R_y=-x. \] Not exact.

Step 2 · Find $\mu=x^ay^bz^c$

Imposing all three exactness conditions on $\mu P,\mu Q,\mu R$ forces $a=0$, $b=-1$, $c=-2$, i.e. \[ \mu = \frac{1}{yz^2}. \]

Step 3 · Multiply through

\[ \frac{1}{z}\,dx - \frac{1}{y}\,dy - \frac{x}{z^2}\,dz = 0, \] which is exact: $\tilde P_y=0=\tilde Q_x$, $\tilde P_z=-1/z^2=\tilde R_x$, $\tilde Q_z=0=\tilde R_y$.

Step 4 · Find the potential $F$

From $F_x=1/z$: $F=x/z+g(y,z)$. From $F_y=-1/y$: $g=-\ln|y|+h(z)$. From $F_z=-x/z^2$: $h'(z)=0$. Hence \[ F = \frac{x}{z}-\ln|y|. \]

Verification

$d\!\left(\dfrac{x}{z}-\ln|y|\right) =\dfrac1z\,dx-\dfrac1y\,dy-\dfrac{x}{z^2}\,dz$. Multiplying by $yz^2$ recovers \[ yz\,dx-z^2\,dy-xy\,dz=0. \checkmark \]