Consider the dynamics of a system governed by the following nonlinear differential equations: d/dt (x) = 15(y - x) d/dt (y) = - y - xz + 20x d/dt (z) = - z + xy
The equilibrium points of the system is/are
a) (-5.23, -5.23, 27) and (5.23, 5.23, 27)
b) (0,0,0)
c) (-7.45, -7.45, 23), (0,0,0) and (7.45, 7.45, 23)
d) (-4.36,-4.36, 19), (0,0,0) and (4.36, 4.36, 19)

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)**.

Consider the dynamics of a system governed by the following nonlinear differential equations: d/dt (x) = 15(y - x) d/dt (y) = - y - xz + 20x d/dt (z) = - z + xy The equilibrium points of the system is/are a) (-5.23, -5.23, 27) and (5.23, 5.23, 27) b) (0,0,0) c) (-7.45, -7.45, 23), (0,0,0) and (7.45, 7.45, 23) d) (-4.36,-4.36, 19), (0,0,0) and (4.36, 4.36, 19)

Explore how to find equilibrium points in a system governed by nonlinear differential equations. This problem involves solving a system of nonlinear differential equations to determine equilibrium conditions. Suitable for advanced undergraduate studies.

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