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

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

Step 1 · Check exactness

With $P=2(y+z)$, $Q=-(x+z)$, $R=2y-x+z$: \[ P_y=2,\ Q_x=-1;\qquad P_z=2,\ R_x=-1;\qquad Q_z=-1,\ R_y=2. \] None match, so the equation is not exact.

Step 2 · Find an integrating factor $\mu(x+z)$

Let $s=x+z$, so $\mu_x=\mu_z=\mu'(s)$. From $(\mu P)_y=(\mu Q)_x$: \[ 2\mu = -\mu's-\mu \implies \frac{\mu'}{\mu}=-\frac{3}{s} \implies \mu=s^{-3}=(x+z)^{-3}. \] The remaining two exactness conditions are then checked directly and also hold, confirming $\mu=(x+z)^{-3}$ is a valid integrating factor.

Step 3 · Find the potential $F$

Multiplying through by $\mu$ and integrating $F_y=-(x+z)^{-2}$ gives $F=-\dfrac{y}{(x+z)^2}+g(x,z)$. Matching $F_x=2(y+z)/(x+z)^3$ forces $g_x=2z/(x+z)^3$, so $g=-\dfrac{z}{(x+z)^2}+h(z)$. Hence \[ F = -\frac{y+z}{(x+z)^2}+h(z). \] Matching $F_z=(2y-x+z)/(x+z)^3$ then forces $h'(z)=0$.

Verification

With $F=-\dfrac{y+z}{(x+z)^2}$: \[ F_x = \frac{2(y+z)}{(x+z)^3},\qquad F_y=-\frac{1}{(x+z)^2},\qquad F_z=\frac{2y-x+z}{(x+z)^3}, \] which are exactly $\mu P$, $\mu Q$, $\mu R$. \checkmark
Note: this problem appeared twice, verbatim, in the source material — the two copies have been merged into this single entry.