Solve the system dx+dy+(x+y) dz=0, z (dx+dy)+(x+y) dz=0
Solve the system \[ dx+dy+(x+y)\,dz=0, \qquad z\,(dx+dy)+(x+y)\,dz=0. \]
Explanation of Method
Both equations depend on $x,y$ only through the combination $u=x+y$; substituting collapses each equation to a simple separable or exact ODE in $u(z)$.
Step 1 · Substitute $u=x+y$
With $du=dx+dy$, the two equations become \[ du+u\,dz=0, \qquad z\,du+u\,dz=0. \]
Step 2 · Solve the first (separable)
\[ \frac{du}{u}=-dz \implies \ln|u|=-z+\text{const} \implies ue^{z}=C_1 \implies (x+y)e^{z}=C_1. \]
Step 3 · Solve the second (already exact)
\[ z\,du+u\,dz = d(uz) = 0 \implies uz=C_2 \implies (x+y)z=C_2. \]
Verification
$d[(x+y)e^z]=e^z[(dx+dy)+(x+y)\,dz]$, which is $e^z$
times the first equation. \checkmark
$d[(x+y)z]=z(dx+dy)+(x+y)\,dz$, which is exactly the second equation.
\checkmark
Together these two independent conditions pin down the
solution curves of the system (the intersection of the two surfaces).
Result
The solution curves of the system are given by: