Solve z dx + (x-y) dy + yz dz = 0

Solve \[ z\,dx + (x-y)\,dy + yz\,dz = 0. \]

Step 1 · Test integrability

With $P=z$, $Q=x-y$, $R=yz$: \[ P(R_y-Q_z)+Q(P_z-R_x)+R(Q_x-P_y) = z^2+x-y+yz \neq 0. \] No integral surface exists; the general solution is a family of curves.

Step 2 · Freeze $z=c$ and solve

On each plane $z=c$ the equation becomes \[ c\,dx+(x-y)\,dy=0 \implies \frac{dx}{dy}+\frac{x}{c}=\frac{y}{c}, \] a linear ODE in $x(y)$. With integrating factor $e^{y/c}$: \[ \frac{d}{dy}\!\left(xe^{y/c}\right)=\frac{y}{c}e^{y/c} \implies xe^{y/c}=e^{y/c}(y-c)+C \implies x=y-c+Ce^{-y/c}. \]

Step 3 · Restore $z$

Replacing $c$ by $z$: \[ x = y-z+Ce^{-y/z}. \]

Verification

On any plane $z=c$ (so $dz=0$), differentiating $x=y-c+Ce^{-y/c}$ gives $dx=\dfrac{y-x}{c}\,dy$, and substituting into $c\,dx+(x-y)\,dy$ gives $(y-x)\,dy+(x-y)\,dy=0$. \checkmark Since the equation is not integrable, this confirms the solution curves lie in the planes $z=\text{const}$, exactly as constructed.