Question

Given that the Laplace transforms of \( J_0(x), J_0'(x), \) and \( J_0''(x) \) exist, where \( J_0(x) \) is the Bessel function. Let \( Y = Y(s) \) be the Laplace transform of the Bessel function \( J_0(x) \). Then, which one of the following is TRUE?

• A. \( \frac{dY}{ds} + \frac{2sY}{s^2 + 1} = 0, \, s>0 \)
• B. \( \frac{dY}{ds} + \frac{2sY}{s^2 + 1} = 0, \, s>0 \)
• C. \( \frac{dY}{ds} - \frac{sY}{s^2 + 1} = 0, \, s>0 \)
• D. \( \frac{dY}{ds} + \frac{sY}{s^2 + 1} = 0, \, s>0 \)

Hint:

When dealing with Laplace transforms of Bessel functions, remember their known properties and standard forms. The Laplace transform of \( J_0(x) \) leads to a differential equation with the structure \( \frac{dY}{ds} + \frac{sY}{s^2 + 1} = 0 \).

Answer 1

The Correct Option is D

For the Bessel function \( J_0(x) \), the Laplace transform \( Y(s) \) satisfies the differential equation:

\[ \frac{dY}{ds} + \frac{sY}{s^2 + 1} = 0, \quad s > 0. \]
This is a standard result for the Laplace transform of \( J_0(x) \).

Now, let’s examine the options:

Option A: \( \frac{dY}{ds} + \frac{2sY}{s^2 + 1} = 0 \)
This is incorrect because the coefficient \( 2s \) is not correct.

Option B: \( \frac{dY}{ds} + \frac{2sY}{s^2 + 1} = 0 \)
This is also incorrect for the same reason as option A.

Option C: \( \frac{dY}{ds} - \frac{sY}{s^2 + 1} = 0 \)
This is incorrect because the sign before the second term is wrong.

Option D: \( \frac{dY}{ds} + \frac{sY}{s^2 + 1} = 0 \)
This is the correct equation, matching the known differential equation for the Laplace transform of \( J_0(x) \).

Thus, Option D is the correct answer.

\[ \boxed{D} \quad \frac{dY}{ds} + \frac{sY}{s^2 + 1} = 0, \, s > 0 \]