Solve by finding first integrals the symmetric system…
Solve by finding first integrals the symmetric system: $$ \frac{dx}{x+y^2+z^2} = \frac{dy}{y} = \frac{dz}{z} $$
Explanation of Method
The last two ratios immediately give a constant ratio $y/z$; substituting that into the first ratio turns it into a first-order linear ODE for $x$ as a function of $z$, solved with an integrating factor.
- Equate the last two ratios to find the first integral $y/z=C_1$.
- Use it to rewrite the first ratio as a linear ODE for $x(z)$.
- Solve with an integrating factor.
- Express the result as the second first integral.
Step 1 · First integral
From $\dfrac{dy}{y}=\dfrac{dz}{z}$, integrating gives
$$ \frac{y}{z}=C_1. $$
Step 2 · Linear ODE for $x(z)$
Using $y=C_1z$ in $\dfrac{dx}{x+y^2+z^2}=\dfrac{dz}{z}$ gives $\dfrac{dx}{dz}=\dfrac{x}{z}+(C_1^2+1)z$, i.e.
$$ \frac{dx}{dz}-\frac{x}{z}=(C_1^2+1)z. $$
Step 3 · Solve with an integrating factor
With $\mu=1/z$, the equation becomes $\dfrac{d}{dz}\!\left(\dfrac{x}{z}\right)=C_1^2+1$, so $\dfrac{x}{z}=(C_1^2+1)z+C_2$, i.e. $x=y^2+z^2+C_2z$.
Step 4 · Second integral
$$ \frac{x-y^2-z^2}{z}=C_2. $$
Step 5 · Verification
Differentiating $y/z$ and $(x-y^2-z^2)/z$ along the flow $\dot x=x+y^2+z^2,\ \dot y=y,\ \dot z=z$ confirms both are conserved. ✓
Summary Table
| Quantity | Value |
|---|---|
| First integral | $y/z=C_1$ |
| Reduced linear ODE | $dx/dz - x/z = (C_1^2+1)z$ |
| Second integral | $(x-y^2-z^2)/z=C_2$ |
Additional Notes
- Once one ratio is eliminated ($y=C_1z$), the remaining relation becomes an ordinary linear first-order ODE — a common simplification in symmetric systems with one "nonlinear-looking" equation.
- Equivalently, the general solution can be written as $x=y^2+z^2+C_2z$ with $y=C_1z$, using $z$ as a free parameter along each curve.
- Verified by direct differentiation of both conserved quantities along the flow.
Result
The first integrals of the system are: