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The given problem is to show that there are no bounded solutions for the differential equation:

\[
y'' + \left( 1 + \frac{1}{1+x^4} \right) y = \cos(x), \quad x \geq 0
\]

### Steps to approach the solution:

#### 1. **Understanding the nature of the differential equation:**
   - The equation is of the form:
     \[
     y'' + p(x)y = f(x)
     \]
     where \( p(x) = 1 + \frac{1}{1+x^4} \) and \( f(x) = \cos(x) \).
     
     The term \( p(x) \) is always greater than or equal to 1 since \( \frac{1}{1+x^4} \geq 0 \) for all \( x \geq 0 \). This indicates that the coefficient of \( y \) is always positive.

#### 2. **Behavior at infinity:**
   - As \( x \to \infty \), \( \frac{1}{1+x^4} \to 0 \), and thus the equation approaches:
     \[
     y'' + y = \cos(x)
     \]
     which is a standard non-homogeneous linear second-order differential equation. The homogeneous equation corresponding to this is:
     \[
     y'' + y = 0
     \]
     which has general solutions of the form:
     \[
     y_h(x) = c_1 \sin(x) + c_2 \cos(x)
     \]
     These are oscillatory solutions that are unbounded as \( x \to \infty \), unless \( c_1 = c_2 = 0 \), which is generally not the case in the presence of a non-homogeneous term.

#### 3. **Particular solution for the non-homogeneous equation:**
   - A particular solution for \( y'' + y = \cos(x) \) can be found by the method of undetermined coefficients. Trying a particular solution of the form \( y_p = A \cos(x) + B \sin(x) \), and substituting into the equation, we find that the particular solution is:
     \[
     y_p(x) = \frac{1}{2} \cos(x)
     \]
     Thus, the general solution to the equation as \( x \to \infty \) is:
     \[
     y(x) = c_1 \sin(x) + c_2 \cos(x) + \frac{1}{2} \cos(x)
     \]
     which is clearly unbounded because the oscillatory terms \( \sin(x) \) and \( \cos(x) \) do not decay to zero as \( x \to \infty \).

#### 4. **Conclusion:**
   - Since the solution contains oscillatory terms that grow without bound as \( x \to \infty \), the solutions are unbounded for large \( x \).
   - The additional term \( \frac{1}{1+x^4} \) only modifies the equation slightly for finite \( x \), but it does not prevent the solution from being unbounded as \( x \to \infty \).

Thus, **there are no bounded solutions** to the given differential equation for \( x \geq 0 \).

---

Would you like more details or have any questions?

Here are 5 related questions:
1. How does the behavior of \( \frac{1}{1+x^4} \) affect the differential equation at large \( x \)?
2. What are the general techniques for solving second-order non-homogeneous differential equations?
3. Why does the presence of oscillatory solutions generally lead to unbounded solutions?
4. How can we determine the particular solution for a differential equation with a trigonometric non-homogeneous term?
5. What is the difference between bounded and unbounded solutions in the context of differential equations?

**Tip:** When solving non-homogeneous linear differential equations, always start by solving the corresponding homogeneous equation first to understand the behavior of the solutions.

Show that there are no bounded solutions for the differential equation y'' + [1 + 1/(1+x^4)]y = cos(x), x ≥ 0.

Explore why the differential equation y'' + [1 + 1/(1+x^4)]y = cos(x) has no bounded solutions for x ≥ 0. This problem involves analyzing oscillatory solutions and the behavior of second-order linear differential equations at infinity. Suitable for advanced mathematics students.

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