2.5.3 Consider the equation x*=rx+x^3, where r > 0 is fixed.
Show that x(t) → ±∞ in finite time, starting from any initial
condition x0 ≠ 0.

To demonstrate that \( x(t) \to \pm \infty \) in finite time for the differential equation

\[
\frac{dx}{dt} = rx + x^3
\]

with \( r > 0 \), and any initial condition \( x_0 \neq 0 \), we proceed by analyzing the behavior of solutions over time.

### Step 1: Analyze the structure of the differential equation
We have a first-order nonlinear differential equation

\[
\frac{dx}{dt} = rx + x^3
\]

Here, \( r > 0 \) is a constant. Notice that the equation contains a linear term \( rx \) and a cubic nonlinearity \( x^3 \).

### Step 2: Separate variables
We will use separation of variables to better understand the time behavior. Rewrite the equation as:

\[
\frac{1}{rx + x^3} \, dx = dt
\]

This allows us to express time \( t \) as an integral of \( x \):

\[
t - t_0 = \int_{x_0}^{x(t)} \frac{1}{rx + x^3} \, dx
\]

### Step 3: Simplify the integrand
Factor the denominator as follows:

\[
rx + x^3 = x(r + x^2)
\]

Thus, the integral becomes:

\[
t - t_0 = \int_{x_0}^{x(t)} \frac{1}{x(r + x^2)} \, dx
\]

This integral becomes difficult to evaluate exactly, but we can analyze its behavior as \( x \to \infty \).

### Step 4: Behavior of the solution for large \( x \)
For large \( x \), the cubic term \( x^3 \) dominates the linear term \( rx \), so for large values of \( x \), the differential equation approximates:

\[
\frac{dx}{dt} \approx x^3
\]

Solving this simpler equation:

\[
\frac{dx}{x^3} = dt
\]

Integrating both sides:

\[
-\frac{1}{2x^2} = t + C
\]

Solving for \( x \), we get:

\[
x(t) = \pm \frac{1}{\sqrt{2(C - t)}}
\]

From this, it's clear that as \( t \to C \), \( x(t) \to \pm \infty \), which shows that \( x(t) \to \pm \infty \) in finite time.

### Step 5: Conclusion
Since the large \( x \) behavior dominates and leads to \( x(t) \to \pm \infty \) in finite time, we conclude that for any initial condition \( x_0 \neq 0 \), the solution will blow up to infinity (either positive or negative) in finite time.

Thus, starting from any initial condition \( x_0 \neq 0 \), the solution \( x(t) \) will tend to \( \pm \infty \) in finite time.

---

Would you like more details or clarification on any specific part?

### Related questions:
1. How does the behavior of the solution change for \( r < 0 \)?
2. Can we calculate the exact time at which \( x(t) \) blows up?
3. How does the initial condition \( x_0 \) influence the time of blow-up?
4. What is the role of the cubic term \( x^3 \) in determining the blow-up time?
5. What happens if \( x_0 = 0 \)?

### Tip:
For equations with cubic nonlinearities, the large \( x \) behavior often dominates, leading to blow-up or other extreme behaviors in finite time. Always check this limit to understand the global behavior of solutions.

2.5.3 Consider the equation x*=rx+x^3, where r > 0 is fixed. Show that x(t) → ±∞ in finite time, starting from any initial condition x0 ≠ 0.

Explore the solution to the differential equation dx/dt = rx + x^3, showing that x(t) → ±∞ in finite time. Learn how to apply separation of variables and analyze nonlinear dynamics. Suitable for undergraduate-level students.

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