Solve yz dx + (xy-yz^3) dy - 2xy dz = 0

Solve \[ yz\,dx + (xy-yz^3)\,dy - 2xy\,dz = 0. \]

Step 1 · Simplify

Assuming $y\neq0$, divide by $y$: \[ z\,dx+(x-z^3)\,dy-2x\,dz=0. \qquad (*) \] Testing integrability, $P(R_y-Q_z)+Q(P_z-R_x)+R(Q_x-P_y)=x\neq0$, so $(*)$ is not integrable to a surface; it defines a family of curves via its characteristic system \[ \frac{dx}{z}=\frac{dy}{x-z^3}=\frac{dz}{-2x}. \]

Step 2 · First integral from $dx$ and $dz$

\[ \frac{dx}{z}=\frac{dz}{-2x} \implies -2x\,dx=z\,dz \implies 2x^2+z^2=c_1. \]

Step 3 · Second integral

Using $2x\,dx=-z\,dz$ from the first integral, \[ dy=\frac{x-z^3}{z}\,dx = \frac{z^3}{2x}\,dz-\frac12\,dz. \] With $z^2=c_1-2x^2$, $\displaystyle\int\frac{z^3}{2x}\,dz=-\int z^2\,dx =-\int(c_1-2x^2)\,dx=-c_1x+\frac{2x^3}{3}$, so \[ y = -c_1x+\frac{2x^3}{3}-\frac{z}{2}+c_2. \] Substituting $c_1=2x^2+z^2$ and simplifying: \[ c_2 = y+\frac{4x^3}{3}+xz^2+\frac{z}{2}. \]

Verification

Both $c_1=2x^2+z^2$ and $c_2=y+\tfrac{4}{3}x^3+xz^2+\tfrac12z$ have gradients orthogonal to $(P,Q,R)=(z,x-z^3,-2x)$: \[ \nabla c_1\cdot(P,Q,R)=4xz+0-2xz\cdot2=0, \] \[ \nabla c_2\cdot(P,Q,R)=(4x^2+z^2)z+(x-z^3)-4x^2z-x=0, \checkmark \] confirming both are genuine first integrals of the characteristic system.