Solve by finding first integrals the symmetric system…
Solve by finding first integrals the symmetric system: $$ \frac{dx}{x} = \frac{dy}{y} = \frac{dz}{z - \sqrt{x^2+y^2+z^2}} $$
Explanation of Method
The first two ratios give an immediate ratio integral; writing $r=\sqrt{x^2+y^2+z^2}$ and computing its rate of change shows $\dot z$ and $\dot r$ are exact opposites, so $z+r$ is conserved.
- Equate the first two ratios to find the first integral.
- Introduce $r=\sqrt{x^2+y^2+z^2}$ and compute its rate of change.
- Compare $\dot z$ and $\dot r$ to find the second integral.
- Verify both integrals and give the explicit parametric solution.
Step 1 · First integral
From $\dfrac{dx}{x}=\dfrac{dy}{y}$, integrating gives
$$ \frac{x}{y}=C_1. $$
Step 2 · Rate of change of $r$
With $r=\sqrt{x^2+y^2+z^2}$, $\dot x=x$, $\dot y=y$, $\dot z=z-r$,
$$ \dot r=\frac{x\dot x+y\dot y+z\dot z}{r}=\frac{x^2+y^2+z(z-r)}{r}=\frac{r^2-zr}{r}=r-z. $$
Step 3 · Second integral
Since $\dot z=z-r$ and $\dot r=r-z$, their sum vanishes: $\dot z+\dot r=0$, so
$$ z+\sqrt{x^2+y^2+z^2}=C_2. $$
Step 4 · Verification
Both $\Phi_1=x/y$ and $\Phi_2=z+r$ have zero total derivative along the flow, confirming they are first integrals. ✓
Summary Table
| Quantity | Value |
|---|---|
| First integral | $x/y=C_1$ |
| Auxiliary relation | $\dot r=r-z$ |
| Second integral | $z+\sqrt{x^2+y^2+z^2}=C_2$ |
Additional Notes
- Introducing the radial distance $r$ is the key step: its rate of change happens to be the exact negative of $\dot z$, which is what makes $z+r$ conserved.
- Geometrically, $\Phi_1=C_1$ is a plane through the $z$-axis and $\Phi_2=C_2$ is a paraboloid-like surface; the solution curves are their intersections.
- Verified by direct differentiation of both quantities along the flow.
Result
The first integrals of the system are: