Problem 99
Easy Separation of Variables

Solve (cos x+e^xy) dx + (e^x+e^yz) dy + e^y dz = 0

Solve \[ (\cos x+e^xy)\,dx + (e^x+e^yz)\,dy + e^y\,dz = 0. \]

Explanation of Method

Check exactness, then integrate successively to build the potential.

Step 1 · Check exactness

With $P=\cos x+e^xy$, $Q=e^x+e^yz$, $R=e^y$: \[ P_y=e^x=Q_x,\qquad P_z=0=R_x,\qquad Q_z=e^y=R_y. \checkmark \] Exact.

Step 2 · Find the potential $F$

From $F_x=\cos x+e^xy$: $F=\sin x+e^xy+g(y,z)$. From $F_y=e^x+e^yz$: $g_y=e^yz\Rightarrow g=ze^y+h(z)$. From $F_z=e^y$: $h'(z)=0$. Hence \[ F = \sin x+ye^x+ze^y. \]


Verification

$F_x=\cos x+ye^x=P$, $F_y=e^x+ze^y=Q$, $F_z=e^y=R$. \checkmark

Result

The general integral is:

\[ \boxed{ \sin x+ye^x+ze^y = C } \]