Questions: Consider the differential equation y'' - (2α - 1)y' + α(α - 1)y = 0 (a) Determine the values of α for which all solutions tend to zero as t → ∞. Interval: (b) Determine the values of α for which all (nonzero) solutions become unbounded as t → ∞. Interval:

Solution Steps

To solve the given differential equation, we need to analyze the characteristic equation associated with it. The characteristic equation will help us determine the behavior of the solutions as \( t \rightarrow \infty \).

1. Write the characteristic equation for the given differential equation.
2. Solve the characteristic equation to find the roots.
3. Analyze the roots to determine the conditions under which the solutions tend to zero or become unbounded as \( t \rightarrow \infty \).

Given the differential equation: \[ y'' - (2\alpha - 1)y' + \alpha(\alpha - 1)y = 0 \] we can write the characteristic equation as: \[ r^2 - (2\alpha - 1)r + \alpha(\alpha - 1) = 0 \]

Solving the characteristic equation, we find the roots: \[ r = \alpha \quad \text{and} \quad r = \alpha - 1 \]

To determine the behavior of the solutions as \( t \to \infty \), we analyze the real parts of the roots.

For all solutions to tend to zero as \( t \to \infty \), both roots must have negative real parts. Therefore, we need: \[ \alpha < 0 \quad \text{and} \quad \alpha - 1 < 0 \] Solving these inequalities, we get: \[ \alpha < 0 \]

For all (nonzero) solutions to become unbounded as \( t \to \infty \), at least one root must have a positive real part. Therefore, we need: \[ \alpha > 0 \quad \text{or} \quad \alpha - 1 > 0 \] Solving these inequalities, we get: \[ \alpha > 1 \]

Final Answer