Solve the symmetric system, subject to the given initial…
Solve the symmetric system, subject to the given initial data: $$ \frac{dx}{x^2-yz} = \frac{dy}{y^2-yz} = \frac{dz}{z(x+y)},\qquad y=1,\ z=-1\ \text{at }x=0 $$
Explanation of Method
Combining the difference of the first two equations with the third gives a clean logarithmic first integral; substituting $w=y/z$ and using that constant turns the remaining ratio into a separable ODE, solved by partial fractions, and the initial data then fixes both constants.
- Combine $d(x-y)$ with the third fraction to find the first integral.
- Substitute $w=y/z$ into the reduced two-variable ratio.
- Separate variables using partial fractions and integrate.
- Apply the initial condition $x=0,\ y=1,\ z=-1$ to fix both constants.
Step 1 · First integral
Since $(x^2-yz)-(y^2-yz)=x^2-y^2=(x-y)(x+y)$, combining with the third fraction gives $\dfrac{d(x-y)}{x-y}=\dfrac{dz}{z}$. Integrating,
$$ \frac{x-y}{z}=C_1. $$
Step 2 · Reduce with $w=y/z$
Using $x=y+C_1z$ in the remaining ratio and substituting $y=wz$ leads, after simplification, to the separable equation
$$ \frac{(2w+C_1)\,dw}{-w(w+1+C_1)}=\frac{dz}{z}. $$
Step 3 · Integrate by partial fractions
Resolving into partial fractions and integrating, then substituting back $w=y/z$ and $C_1=(x-y)/z$, gives the second first integral in implicit form
$$ \Phi_2=\frac{y^{C_1}(x+z)^{2+C_1}}{z^{1+C_1}}=C_2. $$
Step 4 · Apply the initial data
At $x=0,\ y=1,\ z=-1$: from $\Phi_1$, $C_1=(0-1)/(-1)=1$. Substituting into $\Phi_2$ with $C_1=1$ gives $C_2=1\cdot(-1)^3/(-1)^2=-1$.
Step 5 · Particular solution
With $C_1=1$, $\Phi_1$ gives $x-y-z=0$; with $C_2=-1$ and $x=y+z$, $\Phi_2$ reduces to
$$ y(y+2z)^3+z^2=0. $$
Step 6 · Verification
At $x=0,\ y=1,\ z=-1$: $0-1-(-1)=0$ and $1\cdot(1-2)^3+(-1)^2=-1+1=0$, confirming the particular solution satisfies the given initial data. ✓
Summary Table
| Quantity | Value |
|---|---|
| First integral | $(x-y)/z=C_1$ |
| Constants from IC | $C_1=1,\ C_2=-1$ |
| Particular solution | $x-y-z=0,\ \ y(y+2z)^3+z^2=0$ |
Additional Notes
- The second integral is expressed with an exponent that is itself the first integral $C_1$ — unusual notation, but a perfectly legitimate conserved function of $x,y,z$.
- Because $C_1=1$ turns out to be a clean integer for this particular initial data, the general power-law integral collapses to the simple polynomial relation $y(y+2z)^3+z^2=0$.
- The particular solution was checked directly against the given initial values $x=0,\ y=1,\ z=-1$.
Result
The particular solution satisfying the initial data is: