Solve by finding first integrals the symmetric system…
Solve by finding first integrals the symmetric system: $$ \frac{dx}{y(x+y)} = \frac{dy}{-x(x+y)} = \frac{dz}{-(x-y)(2x+2y+z)} $$
Explanation of Method
The first two ratios combine into the exact differential $x\,dx+y\,dy=0$, giving a circular first integral; substituting $s=x+y$ turns the remaining ratio into a linear first-order ODE for $z$ as a function of $s$.
- Equate the first two ratios to get an exact differential in $x,y$.
- Integrate to find the first integral.
- Combine the sum $dx+dy$ with the third ratio and substitute $s=x+y$.
- Solve the resulting linear ODE for $z(s)$ with an integrating factor.
Step 1 · First integral
Equating the first two fractions, $-x(x+y)\,dx=y(x+y)\,dy$, i.e. $x\,dx+y\,dy=0$. Integrating,
$$ x^2+y^2=C_1. $$
Step 2 · Reduce to a two-variable ODE
Adding the first two denominators, $P+Q=-(x-y)(x+y)$, so $\dfrac{d(x+y)}{-(x+y)(x-y)}=d\tau$. Setting $s=x+y$ and combining with the third fraction gives
$$ (2s+z)\,ds = s\,dz. $$
Step 3 · Solve the linear ODE
Rearranged as $\dfrac{dz}{ds}-\dfrac{z}{s}=2$, with integrating factor $\mu=1/s$, this integrates to $\dfrac{z}{s}=2\ln|s|+C_2$, i.e.
$$ \frac{z}{x+y}-2\ln|x+y|=C_2. $$
Step 4 · Verification
Differentiating $x^2+y^2$ and $z/(x+y)-2\ln|x+y|$ along the flow confirms both are conserved identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| First integral | $x^2+y^2=C_1$ |
| Substitution | $s=x+y$ |
| Second integral | $z/(x+y)-2\ln|x+y|=C_2$ |
Additional Notes
- The circular first integral $x^2+y^2=C_1$ arises from the same exact-differential trick used in several other symmetric-system problems in this set.
- The logarithmic term in the second integral appears because the reduced linear ODE for $z(s)$ has a resonant right-hand side proportional to $1/s$.
- Verified by direct differentiation of both conserved quantities along the flow.
Result
The first integrals of the system are: