Solve the system of differential equations: (dx)/(1) =…
Solve the system of differential equations: $$ \frac{dx}{1} = \frac{dy}{y+z} = \frac{dz}{x+y+z} $$
Explanation of Method
For a symmetric (Lagrange–Charpit) system, forming the combination $d(z-y)$ eliminates the coupling and integrates directly; a second integral follows by solving a linear first-order ODE for $y+z$ in terms of $x$.
- Compute $d(z-y)$ using the common ratio to find a first integral.
- Use that integral to write a linear ODE for $y+z$ as a function of $x$.
- Solve the linear ODE with an integrating factor.
- Express the two independent first integrals.
Step 1 · First integral
With $d\tau=dx$ the common parameter, $d(z-y)=dz-dy=x\,d\tau=x\,dx$. Integrating,
$$ 2(z-y)-x^2=C_1. $$
Step 2 · Linear ODE for $y+z$
Substituting $z=y+\tfrac{x^2}{2}+\tfrac{C_1}{2}$ into $\dfrac{dy}{dx}=y+z$ gives
$$ \frac{dy}{dx}-2y=\frac{x^2}{2}+\frac{C_1}{2}. $$
Step 3 · Solve with an integrating factor
Using $e^{-2x}$ as the integrating factor and integrating by parts, then converting back to $y+z$, gives the second first integral
$$ \left(y+z+\frac x2+\frac14\right)e^{-2x}=C_2. $$
Step 4 · Verification
Differentiating $\Phi_1=2(z-y)-x^2$ and $\Phi_2=(y+z+\tfrac x2+\tfrac14)e^{-2x}$ along the system shows both are conserved: $d\Phi_1=0$ and $d\Phi_2=0$. ✓
Summary Table
| Quantity | Value |
|---|---|
| First integral | $2(z-y)-x^2=C_1$ |
| Second integral | $(y+z+\tfrac x2+\tfrac14)e^{-2x}=C_2$ |
| Solution family | intersection of the two level surfaces $\Phi_1=C_1$, $\Phi_2=C_2$ |
Additional Notes
- A first-order symmetric system of two independent equations has exactly two independent first integrals; any solution curve is the intersection of the two corresponding surfaces.
- The general solution is any functional relation $F(\Phi_1,\Phi_2)=0$, not just an explicit formula.
- Both integrals were verified by showing their total derivative along the flow vanishes identically.
Result
The general solution of the system is: