We are given the differential equation

$\frac{dy}{dx} = (y^2 - 5y + 4) \sin^2 \left( \frac{2\pi y}{7} \right)$

with the initial condition $y(0) = a$, and we are asked to find a possible value of $a$ such that the solution to this problem is a non-constant function that satisfies

$\lim_{x \to \infty} y(x) = \frac{7}{2}.$

Step 1: Identify Equilibrium Points

The right-hand side of the differential equation is given by

$f(y) = (y^2 - 5y + 4) \sin^2 \left( \frac{2\pi y}{7} \right).$

For equilibrium points, we need to solve for values of $y$ such that $f(y) = 0$. This can happen if either:

1. $y^2 - 5y + 4 = 0$, or
2. $\sin^2 \left( \frac{2\pi y}{7} \right) = 0$.

Case 1: $y^2 - 5y + 4 = 0$

Solve the quadratic equation:

$y^2 - 5y + 4 = (y - 1)(y - 4) = 0.$

Thus, $y = 1$ or $y = 4$ are equilibrium points.

Case 2: $\sin^2 \left( \frac{2\pi y}{7} \right) = 0$

This implies $\sin \left( \frac{2\pi y}{7} \right) = 0$, which occurs when

$\frac{2\pi y}{7} = n\pi, \quad n \in \mathbb{Z}.$

Thus, $y = \frac{7n}{2}$. The values of $y$ that satisfy this equation include $y = 0, \frac{7}{2}, 7, \dots$.

Step 2: Behavior of the Solution

We know that the solution satisfies $\lim_{x \to \infty} y(x) = \frac{7}{2}$, which suggests that $y = \frac{7}{2}$ is an attractive equilibrium point for the solution.

Step 3: Analyze the Initial Condition

To ensure that the solution tends toward $y = \frac{7}{2}$ as $x \to \infty$, we need to choose the initial value $a$ such that the solution starts near $y = \frac{7}{2}$ but does not settle at any other equilibrium point (such as $y = 1$ or $y = 4$).

Since $y = 1$ and $y = 4$ are also equilibrium points, and we want the solution to approach $\frac{7}{2}$ as $x \to \infty$, a reasonable choice for $a$ is a value close to, but not equal to, $\frac{7}{2}$.

Step 4: Conclusion

A possible value of $a$ that would lead to a non-constant solution tending to $y = \frac{7}{2}$ is $a = 4$. This is because starting near $y = 4$ allows the solution to move towards the attractive equilibrium point $\frac{7}{2}$ without remaining at $y = 4$ (as $y = 4$ is an unstable equilibrium for this problem).

Final Answer:

A possible value of $a$ is $a = 4$.

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Would you like more details or have any further questions?

Related Questions:

1. How do we analyze the stability of equilibrium points in a differential equation?
2. What is the significance of the function $\sin^2 \left( \frac{2\pi y}{7} \right)$ in this context?
3. Could $a = 1$ also lead to a solution that tends towards $y = \frac{7}{2}$?
4. How would the solution behave if $a = 0$?
5. Can we have a solution where $y(x)$ oscillates between equilibrium points?

Tip:

In problems involving equilibrium points, analyzing their stability is key to understanding the long-term behavior of the solutions.