Problem 92
Easy Separation of Variables

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

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

Explanation of Method

Check the three exactness conditions directly; since they all hold, integrate successively to build the potential $F(x,y,z)$.

Step 1 · Check exactness

With $P=2y-z$, $Q=2x-2z$, $R=-(x+2y)$: \[ P_y=2=Q_x,\qquad P_z=-1=R_x,\qquad Q_z=-2=R_y. \checkmark \] The equation is exact.

Step 2 · Find the potential $F$

From $F_x=2y-z$: $F=2xy-xz+g(y,z)$. From $F_y=2x-2z$: $g_y=-2z\Rightarrow g=-2yz+h(z)$. From $F_z=-(x+2y)$: $h'(z)=0$. Hence \[ F = 2xy-xz-2yz. \]


Verification

$F_x=2y-z=P$, $F_y=2x-2z=Q$, $F_z=-x-2y=R$. \checkmark

Result

The general integral is:

\[ \boxed{ 2xy-xz-2yz = C } \]