Solve the system yz dx+xz dy+xy dz=0, z^2 dx+z^2…
Solve the system \[ yz\,dx+xz\,dy+xy\,dz=0, \qquad z^2\,dx+z^2\,dy+(xz+yz-xy)\,dz=0. \]
Explanation of Method
The first equation is exact and gives one first integral directly. The second is not exact on its own, but substituting the relation from the first integral reduces it to a linear ODE in $s=x+y$.
Step 1 · Solve the first equation
With $P_1=yz$, $Q_1=xz$, $R_1=xy$, all three exactness conditions hold, giving \[ u_1 = xyz = C_1. \]
Step 2 · Reduce the second equation
With $P_2=Q_2=z^2$, $R_2=xz+yz-xy$: $(P_2)_z=2z\neq(R_2)_x=z-y$, not exact. Using $xy=C_1/z$ from the first integral and setting $s=x+y$ (so $ds=dx+dy$), the second equation becomes \[ z^2\,ds+\left(sz-\frac{C_1}{z}\right)dz = 0 \implies ds+\frac{s}{z}\,dz = \frac{C_1}{z^3}\,dz, \] a linear ODE in $s(z)$ with integrating factor $z$: \[ d(sz) = \frac{C_1}{z^2}\,dz \implies sz = -\frac{C_1}{z}+C_2. \]
Step 3 · Restore $s=x+y$, $C_1=xyz$
\[ (x+y)z = -xy+C_2 \implies xy+xz+yz = C_2. \]
Verification
$u_1=xyz$: $(u_1)_x=yz=P_1$, $(u_1)_y=xz=Q_1$,
$(u_1)_z=xy=R_1$. \checkmark
For $u_2=xy+xz+yz$: on the surface $xyz=C_1$, $d(xy)=-\tfrac{xy}{z}\,dz$, so
dividing the second equation by $z$ gives exactly
$z(dx+dy)+(x+y-xy/z)\,dz = d[(x+y)z+xy]=0$. \checkmark
Result
The solution curves of the system are given by: