Solve the system (y+z) dx+(z+x) dy+(x+y) dz=0, (x+z) dx+y…
Solve the system \[ (y+z)\,dx+(z+x)\,dy+(x+y)\,dz=0, \qquad (x+z)\,dx+y\,dy+x\,dz=0. \]
Explanation of Method
Since each equation is separately exact, solve each for its own potential; the two potentials give two independent first integrals whose intersection is the solution curve of the system.
Step 1 · Solve the first equation
With $P_1=y+z$, $Q_1=z+x$, $R_1=x+y$: $(P_1)_y=1=(Q_1)_x$, $(P_1)_z=1=(R_1)_x$, $(Q_1)_z=1=(R_1)_y$. \checkmark Exact. Integrating gives \[ u_1 = xy+xz+yz = C_1. \]
Step 2 · Solve the second equation
With $P_2=x+z$, $Q_2=y$, $R_2=x$: $(P_2)_y=0=(Q_2)_x$, $(P_2)_z=1=(R_2)_x$, $(Q_2)_z=0=(R_2)_y$. \checkmark Exact. Integrating gives \[ u_2 = \frac{x^2}{2}+xz+\frac{y^2}{2} = C_2,\quad\text{i.e.}\quad x^2+y^2+2xz = C_2'. \]
Verification
$(u_1)_x=y+z=P_1$, $(u_1)_y=x+z=Q_1$, $(u_1)_z=x+y=R_1$; $(u_2)_x=x+z=P_2$, $(u_2)_y=y=Q_2$, $(u_2)_z=x=R_2$. \checkmark
Result
The solution curves of the system are the intersections given by: