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

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

Step 1 · Substitute $u=x^3$

With $du=3x^2\,dx$, the equation becomes \[ (y+z)\,du+(z^2-u)\,dy+(y^2-u)\,dz = 0. \qquad (*) \]

Step 2 · Check integrability of $(*)$

With $\tilde P=y+z$, $\tilde Q=z^2-u$, $\tilde R=y^2-u$: \[ \tilde P(\tilde R_y-\tilde Q_z)+\tilde Q(\tilde P_z-\tilde R_u)+\tilde R(\tilde Q_u-\tilde P_y) = 2(y^2-z^2)+2(z^2-u)-2(y^2-u) = 0. \checkmark \]

Step 3 · Find the integrating factor $\mu=(y+z)^{-2}$ and the potential $G$

The three exactness conditions for $\mu(*)$ hold with $\mu=(y+z)^{-2}$. Integrating $G_u=1/(y+z)$ gives $G=u/(y+z)+g(y,z)$; matching $G_y$ forces $g=-z^2/(y+z)+h(z)$; matching $G_z$ forces $h'(z)=1$. Hence \[ G = \frac{u-z^2}{y+z}+z. \]

Step 4 · Restore $u=x^3$

\[ \frac{x^3-z^2}{y+z}+z = C \implies x^3+yz = C(y+z). \]

Verification

Let $F=(x^3+yz)/(y+z)$: \[ F_x=\frac{3x^2}{y+z},\quad F_y=\frac{z^2-x^3}{(y+z)^2},\quad F_z=\frac{y^2-x^3}{(y+z)^2}. \] Multiplying $dF=0$ by $(y+z)^2$ recovers the original equation exactly. \checkmark