Problem 19
Easy Elimination Methods

Solve the system of differential equations: dot x = y - z…

Solve the system of differential equations: $$ \dot x = y - z,\qquad \dot y = x^2 + y,\qquad \dot z = x^2 + z $$

Explanation of Method

Subtracting the last two equations shows $y-z$ satisfies a simple linear relation with $x$, giving a clean first integral $y-z-x=C_1$; substituting this into a combination of the equations reduces the remaining freedom to a first-order linear ODE, solved with an integrating factor.

  1. Subtract the second and third equations to relate $y-z$ to $x$.
  2. Combine with the first equation to get the first integral.
  3. Use it to write a linear ODE for $y+z$ as a function of $x$.
  4. Solve with an integrating factor to get the second integral.

Step 1 · First integral

Subtracting, $\dot y-\dot z=y-z$; combined with $\dot x=y-z$ this gives $d(y-z)=dx$, so

$$ y-z-x=C_1. $$

Step 2 · Linear ODE for $y+z$

Writing $p=y+z$ and using $y-z=x+C_1$, the ratio $\dfrac{d(y+z)}{2x^2+y+z}=\dfrac{dx}{x+C_1}$ becomes the linear ODE

$$ \frac{dp}{dx}-\frac{p}{x+C_1}=\frac{2x^2}{x+C_1}. $$

Step 3 · Solve with an integrating factor

Using $\mu=1/(x+C_1)$ and integrating, the second first integral is

$$ \frac{y+z}{y-z}-2x+4(y-z-x)\ln|y-z|+\frac{2(y-z-x)^2}{y-z}=C_2. $$

Step 4 · Verification

Differentiating $\Phi_1=y-z-x$ along the system gives $\dot\Phi_1=(y-z)-(y-z)=0$, confirming the first integral. The special case $C_1=0$ (i.e. $y-z=x$) reduces to the explicit solution $y+z=2x^2+C_2x$, verified directly. ✓


Summary Table

Quantity Value
First integral$y-z-x=C_1$
Special case $C_1=0$$y+z=2x^2+C_2x$
Second integral (general)involves $\ln|y-z|$

Additional Notes

Result

The general solution of the system is:

$$ \begin{aligned} &y - z - x = C_1,\\ &\frac{y+z}{y-z} - 2x + 4C_1\ln|y-z| + \frac{2C_1^2}{y-z} = C_2. \end{aligned} $$