Question

Let \( P : \mathbb{R} \to \mathbb{R} \) be a continuous function such that \( P(x)>0 \) for all \(x \in \mathbb{R}\). Let \(y\) be a twice differentiable function on \(\mathbb{R}\) satisfying \[ y''(x) + P(x)y'(x) - y(x) = 0 \] for all \(x \in \mathbb{R}\). Suppose that there exist two real numbers \(a,b\) (\(a<b\)) such that \(y(a) = y(b) = 0\). Then

• A. \(y(x) = 0\) for all \(x \in [a,b]\).
• B. \(y(x)>0\) for all \(x \in (a,b)\).
• C. \(y(x)<0\) for all \(x \in (a,b)\).
• D. \(y(x)\) changes sign on \((a,b)\).

Hint:

For second-order ODEs of the form \(y'' + P(x)y' - y = 0\) with \(P(x)>0\), any non-trivial solution that is zero at two distinct points must change sign between them.

Answer 1

The Correct Option is A

Step 1: Analyze the differential equation.
Multiply both sides by \( e^{\int P(x)\,dx} \): \[ \frac{d}{dx}\big(y'(x)e^{\int P(x)\,dx}\big) = y(x)e^{\int P(x)\,dx}. \] Integrating from \(a\) to \(b\), \[ y'(b)e^{\int_a^b P(x)\,dx} - y'(a) = \int_a^b y(x)e^{\int_a^x P(t)\,dt} \, dx. \]
Step 2: Applying boundary conditions.
Since \(y(a)=y(b)=0\), if \(y(x)\) does not change sign on \((a,b)\), the right-hand side integral would have a fixed sign. This implies \(y'(a)\) and \(y'(b)\) must have the same sign — which is impossible for \(y\) to return to zero at both ends.
Step 3: Conclusion.
Hence, \(y(x)\) must change sign in \((a,b)\).