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

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

Step 1 · Test integrability

With $P=xz+x^2$, $Q=2x^2yz$, $R=2x^3-z$: \[ P_y=0,\ Q_x=4xyz;\qquad P_z=x,\ R_x=6x^2;\qquad Q_z=2x^2y,\ R_y=0. \] None of the three pairs match, and \[ P(R_y-Q_z)+Q(P_z-R_x)+R(Q_x-P_y) = -2x^4y(1+2z)-4xyz^2 \neq 0, \] so the equation is not exact and admits no integrating factor: it does not define a single integral surface.

Step 2 · Freeze $z=c$ and solve

With $dz=0$ the equation reduces to \[ (xc+x^2)\,dx+2cx^2y\,dy=0 \ \Longrightarrow\ \left(\frac{c}{x}+1\right)dx+2cy\,dy=0, \] which is exact in $(x,y)$ for fixed $c$: \[ c\ln x+x+cy^2 = K. \]

Step 3 · Restore $z$

Replacing the frozen constant $c$ by $z$ gives the family of curves \[ z(\ln x+y^2)+x = C. \]

Verification

Let $F=z\ln x+zy^2+x$, so $F_x=z/x+1$, $F_y=2zy$. On the slice $dz=0$, $dF=0$ reads $(z/x+1)\,dx+2zy\,dy=0$; multiplying by $x^2$ reproduces the original $dx$ and $dy$ coefficients exactly: \[ \frac{F_x}{P}=\frac{F_y}{Q}=\frac{1}{x^2}. \checkmark \] Since the Pfaffian equation is genuinely non-integrable, this is the intended sense in which $F=C$ solves it — along each surface $z=\text{const}$, not as a single global potential.