Explain why the graph of the solution to the initial value problem y'(t) = t²/(1 - t), y(-1) = ln 2 cannot cross the line t = 1.

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Identify the differential equation given: \(y'(t) = \frac{t^2}{1 - t}\).

Notice that the denominator \(1 - t\) becomes zero at \(t = 1\), which means the right-hand side of the differential equation is undefined at \(t = 1\).

Since the derivative \(y'(t)\) is not defined at \(t = 1\), the solution \(y(t)\) cannot be extended continuously through \(t = 1\); this creates a vertical asymptote or a discontinuity in the slope.

The initial condition \(y(-1) = \ln 2\) is given at \(t = -1\), which is less than 1, so the solution is defined on an interval containing \(-1\) but cannot cross the point where the derivative is undefined, \(t = 1\).

Therefore, the graph of the solution cannot cross the vertical line \(t = 1\) because the differential equation does not allow the solution to be differentiable or continuous at that point.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Existence and Uniqueness Theorem

This theorem states that if the function defining the differential equation and its partial derivative with respect to y are continuous near the initial condition, then a unique solution exists locally. It ensures the solution behaves predictably and helps identify points where the solution may fail to extend.

Domain of the Differential Equation

The domain is the set of t-values where the function defining y' is defined and continuous. Since y'(t) = t²/(1 - t) has a denominator that becomes zero at t = 1, the function is undefined there, creating a vertical asymptote or discontinuity that the solution cannot cross.

Behavior Near Singularities

When the differential equation has singularities (points where it is undefined), solutions often approach these points asymptotically but cannot cross them. At t = 1, the denominator zero causes the slope to become unbounded, preventing the solution curve from crossing this vertical line.