Problem 98
Solve (y+3z) dx + (x+2z) dy + (3x+2y) dz = 0
Solve \[ (y+3z)\,dx + (x+2z)\,dy + (3x+2y)\,dz = 0. \]
Explanation of Method
Check the three exactness conditions; since they all hold, integrate successively to build the potential $F(x,y,z)$.
Step 1 · Check exactness
With $P=y+3z$, $Q=x+2z$, $R=3x+2y$: \[ P_y=1=Q_x,\qquad P_z=3=R_x,\qquad Q_z=2=R_y. \checkmark \] The equation is exact.
Step 2 · Find the potential $F$
From $F_x=y+3z$: $F=xy+3xz+g(y,z)$. From $F_y=x+2z$: $g_y=2z\Rightarrow g=2yz+h(z)$. From $F_z=3x+2y$: $h'(z)=0$. Hence \[ F = xy+3xz+2yz. \]
Verification
$F_x=y+3z=P$, $F_y=x+2z=Q$, $F_z=3x+2y=R$. \checkmark
Result
The general integral is:
\[ \boxed{ xy+3xz+2yz = C } \]