To find the equilibrium points of the system, we need to solve the system of nonlinear differential equations by setting the time derivatives to zero. The system of equations is:

$\frac{dx}{dt} = 15(y - x)$ $\frac{dy}{dt} = -y - xz + 20x$ $\frac{dz}{dt} = -z + xy$

Step 1: Set the derivatives to zero

At equilibrium, $\frac{dx}{dt} = \frac{dy}{dt} = \frac{dz}{dt} = 0$. So:

1. $15(y - x) = 0$
2. $-y - xz + 20x = 0$
3. $-z + xy = 0$

Step 2: Solve the equations

1. From the first equation:

$y - x = 0 \quad \Rightarrow \quad y = x$

2. Substitute $y = x$ into the second equation:

$-x - xz + 20x = 0$ $x(-1 - z + 20) = 0 \quad \Rightarrow \quad x(-z + 19) = 0$

This gives us two possible cases:

• $x = 0$
• $z = 19$

3. Consider the third equation with $y = x$:

$-z + xy = 0 \quad \Rightarrow \quad z = xy = x^2$

Step 3: Analyze the cases

Case 1: $x = 0$

• If $x = 0$, then $y = 0$, and substituting into $z = x^2$ gives $z = 0$.
• So, one equilibrium point is $(0, 0, 0)$.

Case 2: $z = 19$

• If $z = 19$, then $z = x^2$ implies $x^2 = 19$.
• Thus, $x = \sqrt{19} \approx \pm 4.36$.
• Since $y = x$, the equilibrium points in this case are $(4.36, 4.36, 19)$ and $(-4.36, -4.36, 19)$.

Conclusion

The equilibrium points of the system are $(0, 0, 0)$, $(4.36, 4.36, 19)$, and $(-4.36, -4.36, 19)$, which corresponds to option d).

So, the correct answer is:

d) (-4.36,-4.36, 19), (0,0,0) and (4.36, 4.36, 19).