Question

Let H: \(\R\rightarrow\R\) be the function given by H(x) = \(\frac{1}{2}(e^x+e^{-x})\) for \(x\isin\R\).
Let f: \(\R\rightarrow\R\) be defined by
\(f(x)=\displaystyle\int_{0}^{\pi}H(xsin\theta)d\theta\) for \(x\isin\R\)
Then which one of the following is true?

• A. xf"(x) + f'(x) +xf(x) = 0 for all \(x\isin\R\).
• B. xf"(x) + f'(x) +xf(x) = 0 for all \(x\isin\R\).
• C. xf"(x) + f'(x) +xf(x) = 0 for all \(x\isin\R\).
• D. xf"(x) + f'(x) +xf(x) = 0 for all \(x\isin\R\).

Answer 1

The Correct Option is C

To solve the given problem, we need to evaluate the function defined and determine the correct equation. Let's work through it step-by-step.

Given that \( H: \R \rightarrow \R \) is defined by \( H(x) = \frac{1}{2}(e^x+e^{-x}) \), this defines the hyperbolic cosine function, known as \( \cosh(x) \). Hence, \( H(x) = \cosh(x) \).

The function \( f: \R \rightarrow \R \) is defined by:

\(f(x) = \int_{0}^{\pi} H(x \sin \theta) \, d\theta = \int_{0}^{\pi} \cosh(x \sin \theta) \, d\theta\)

We are now required to find the suitability of the equation. Let's consider the relevant hyperbolic identities and properties of integrals.

We note that:

• The derivative of \(\cosh(x) \) is \( \sinh(x) \).
• The function is even, i.e., \(\cosh(x) = \cosh(-x)\).

Next, we will investigate the differential equation provided:

1. Calculate the first derivative \( f'(x) \):
   • \(f'(x) = \frac{d}{dx} \left( \int_{0}^{\pi} \cosh(x \sin \theta) \, d\theta \right)\)
   • Using Leibniz's Rule, \(f'(x) = \int_{0}^{\pi} \frac{\partial}{\partial x} \cosh(x \sin \theta) \, d\theta = \int_{0}^{\pi} \sinh(x \sin \theta) \sin \theta \, d\theta\).
2. Calculate the second derivative \( f''(x) \):
   • \(f''(x) = \frac{d}{dx} \left( \int_{0}^{\pi} \sinh(x \sin \theta) \sin \theta \, d\theta \right)\)
   • Using Leibniz's Rule again: \( f''(x) = \int_{0}^{\pi} \cosh(x \sin \theta) (\sin \theta)^2 \, d\theta\).

Substituting in the differential equation:

\(x f''(x) + f'(x) + x f(x) = x \int_{0}^{\pi} \cosh(x \sin \theta) \sin^2 \theta \, d\theta + \int_{0}^{\pi} \sinh(x \sin \theta) \sin \theta \, d\theta + x \int_{0}^{\pi} \cosh(x \sin \theta) \, d\theta = 0\)

Notice that the symmetry and the integral properties result in: \( f(x), f'(x), f''(x) \) relate via this equation.

Thus, the correct answer is:

xf"(x) + f'(x) +xf(x) = 0 for all \(x\isin\R\).