Solve the system dx+2 dy-(x+2y) dz=0, 2 dx+dy+(x-y) dz=0
Solve the system \[ dx+2\,dy-(x+2y)\,dz=0, \qquad 2\,dx+dy+(x-y)\,dz=0. \]
Explanation of Method
Treat $z$ as the independent variable and $x,y$ as unknowns; solve the resulting linear algebraic system for $x'=dx/dz$ and $y'=dy/dz$, then integrate the two decoupled/triangular linear ODEs.
Step 1 · Set up and solve for $x',y'$
Dividing by $dz$: \[ x'+2y'=x+2y, \qquad 2x'+y'=y-x. \] Taking $2\times(1)-(2)$ gives $3y'=3x+3y\Rightarrow y'=x+y$. Taking $2\times(2)-(1)$ gives $3x'=-3x\Rightarrow x'=-x$.
Step 2 · Solve $x'=-x$
\[ x = C_1e^{-z} \implies u_1 = xe^{z} = C_1. \]
Step 3 · Solve $y'-y=x=C_1e^{-z}$
Linear ODE with integrating factor $e^{-z}$: \[ \frac{d}{dz}(ye^{-z}) = C_1e^{-2z} \implies ye^{-z}=-\frac{C_1}{2}e^{-2z}+C_2. \] Multiplying by $e^z$ and substituting $C_1=xe^z$: \[ y = -\frac{x}{2}+C_2e^{z} \implies (x+2y)e^{-z} = 2C_2. \]
Verification
Along solutions $x'=-x$, $y'=x+y$: \[ \frac{d}{dz}(xe^z) = e^z(x'+x) = 0. \checkmark \] \[ \frac{d}{dz}\bigl[(x+2y)e^{-z}\bigr] = e^{-z}\bigl[x'+2y'-(x+2y)\bigr] = e^{-z}\bigl[-x+2(x+y)-x-2y\bigr] = 0. \checkmark \]
Result
The solution curves of the system are given by: