Solve the system 2yz dx+x(z dy+y dz)=0, y dx-x^2z dy+y dz=0

Solve the system \[ 2yz\,dx+x(z\,dy+y\,dz)=0, \qquad y\,dx-x^2z\,dy+y\,dz=0. \]

Step 1 · Solve the first equation

Expanded: $2yz\,dx+xz\,dy+xy\,dz=0$. With $P_1=2yz$, $Q_1=xz$: $(P_1)_y=2z\neq(Q_1)_x=z$, not exact. Trying $\mu=\mu(x)$: $(\mu P_1)_y=(\mu Q_1)_x$ gives $2\mu z=\mu z+\mu'xz \Rightarrow \mu'/\mu=1/x \Rightarrow \mu=x$ (the remaining two conditions also check out). Multiplying by $x$ and integrating: \[ u_1 = x^2yz = C_1. \]

Step 2 · Reduce the second equation

With $P_2=y$, $Q_2=-x^2z$, $R_2=y$: $(P_2)_y=1\neq(Q_2)_x=-2xz$, not exact. Using $x^2z=C_1/y$ from the first integral: \[ y\,dx-\frac{C_1}{y}\,dy+y\,dz=0 \implies dx+dz-\frac{C_1}{y^2}\,dy=0. \] Since $d(C_1/y)=-C_1\,dy/y^2$, this is exactly \[ dx+dz+d\!\left(\frac{C_1}{y}\right)=0 \implies x+z+\frac{C_1}{y}=C_2. \]

Step 3 · Restore $C_1=x^2yz$

\[ x+z+\frac{x^2yz}{y} = C_2 \implies x+z+x^2z = C_2. \]

Verification

$u_1=x^2yz$: $(u_1)_x=2xyz$, $(u_1)_y=x^2z$, $(u_1)_z=x^2y$ — exactly $x$ times the expanded first equation's coefficients $(2yz,xz,xy)$. \checkmark
$u_2=x+z+x^2z$: $(u_2)_x=1+2xz$, $(u_2)_y=0$, $(u_2)_z=1+x^2$; on the constraint surface $x^2z=C_1/y$ this differential matches the second equation after the substitution used to derive it. \checkmark