Question

Let S be the set of all real numbers α such that the solution y of the initial value problem
\(\frac{dy}{dx}=y(2-y),\\y(0)=\alpha,\)
exists on [0, ∞). Then the minimum of the set S is equal to __________. (rounded off to two decimal places)

Answer 1

Correct Answer: -0.01 - 0.01

The given differential equation is \( \frac{dy}{dx} = y(2 - y) \) with the initial condition \( y(0) = \alpha \). First, we analyze the behavior of the solutions to determine the set \( S \) of all possible \( \alpha \) for which the solution exists on \( [0, \infty) \).

The equation can be rewritten as \( \frac{dy}{y(2 - y)} = dx \). Separating variables and integrating both sides, we have:

\(\int \frac{1}{y(2 - y)} \, dy = \int dx.\)

To integrate \( \frac{1}{y(2 - y)} \), we use partial fraction decomposition:

\(\frac{1}{y(2 - y)} = \frac{A}{y} + \frac{B}{2 - y}.\)

Solving for \( A \) and \( B \), we get:

\(A = \frac{1}{2}, \, B = -\frac{1}{2}.\)

Integrating, we have:

\(\int \left( \frac{1}{2y} - \frac{1}{2(2 - y)} \right) \, dy = \int dx\)

Which simplifies to:

\(\Rightarrow \frac{1}{2} \ln|y| - \frac{1}{2} \ln|2 - y| = x + C.\)

Exponentiating both sides, we obtain:

\(\frac{|y|}{|2 - y|} = C e^{2x}.\)

Applying the initial condition \( y(0) = \alpha \), we find:

\(\frac{|\alpha|}{|2 - \alpha|} = C.\)

The solution on \( [0, \infty) \) means we need \( y \to 2 \) as \( x \to \infty \) without reaching or exceeding this value in finite time. Analyzing the phase portrait:

• If \( y(0) = 0 \), then \( y = 0 \);
• If \( y(0) = 2 \), then the solution is constant, \( y = 2 \).

Since \( \frac{dy}{dx} = y(2 - y) \), solutions stay within \( (0, 2) \) if the initial value lies therein. The minimum possible \( \alpha \) is thus slightly above zero to ensure that \( y \) cannot remain at zero due to potential perturbations.

Conclude by confirming with the given range:

The minimum of \( S \) is \( 0 \), matching the range \( -0.01, 0.01 \) when rounded to two decimal places.

Final Answer: The minimum value of \( S \) is 0.00.