Solve 2x(y+z) dx + (2yz-x^2+y^2-z^2) dy + (2yz-x^2-y^2+z^2)…

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

Step 1 · Check exactness and integrability

With $P=2x(y+z)$, $Q=2yz-x^2+y^2-z^2$, $R=2yz-x^2-y^2+z^2$: $P_y=2x\neq Q_x=-2x$, not exact, but \[ P(R_y-Q_z)+Q(P_z-R_x)+R(Q_x-P_y) = 8x(z^2-y^2)+8x(y^2-z^2) = 0, \] so an integrating factor exists.

Step 2 · Find $\mu=\mu(y+z)$

Let $s=y+z$. From $(\mu P)_y=(\mu Q)_x$: \[ 2\mu's+4x\mu\cdot\frac{1}{2x}\cdot2 = 0\ \Longrightarrow\ \mu's+2\mu=0 \implies \mu=(y+z)^{-2}. \] The remaining two exactness conditions are checked directly and also hold.

Step 3 · Find the potential $F$

From $F_x=2x/(y+z)$: $F=x^2/(y+z)+g(y,z)$. Matching $F_y$ gives $g_y=1-2z^2/(y+z)^2$, so $g=y+2z^2/(y+z)+h(z)$; matching $F_z$ then gives $h'(z)=-1$. Hence \[ F = \frac{x^2+2z^2}{y+z}+y-z. \] Multiplying by $(y+z)$ and simplifying ($2z^2+(y+z)(y-z)=y^2+z^2$): \[ x^2+y^2+z^2 = C(y+z). \]

Verification

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