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Let $P : \mathbb{R} \rightarrow \mathbb{R}$ 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 $(a, b)$.
in MA2021 by Professor | 120 views

1 Answer

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The correct answer is:

(D) $y(x)$ changes sign on $(a, b)$.


Given that $y(a) = y(b) = 0$, and $y(x)$ is continuous, this implies that there exists at least one point $c \in (a, b)$ such that $y(c) = 0$. This is because $y(x)$ is continuous on $[a, b]$ and satisfies $y(a) = y(b) = 0$, so it must cross the $x$-axis at least once in the interval $(a, b)$.

Now, we need to determine whether $y(x)$ changes sign on $(a, b)$ or not. For this, we can use the Intermediate Value Theorem. Since $y(x)$ is continuous and $y(c) = 0$ for some $c \in (a, b)$, there must be some point $d \in (a, c)$ and some point $e \in (c, b)$ such that $y(d) < 0$ and $y(e) > 0$, respectively. This is because $y(x)$ must pass through all values between negative and positive as it crosses the $x$-axis.

Hence, option (D) is correct: $y(x)$ changes sign on $(a, b)$.

This answer is AI generated but verified by expert.

by Professor
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