Question

Let \(f\) be a differential function with
\[ \lim_{x \to \infty} f(x) = 0. \text{ If } y' + y f'(x) - f(x) f'(x) = 0, \lim_{x \to \infty} y(x) = 0 \text{ then,} \]

• A. \(y + 1 = e^{-f(x)} + f(x)\)
• B. \(y + 1 = e^{-f(x)} + f(x)\)
• C. \(y + 2 = e^{-f(x)} + f(x)\)
• D. \(y - 1 = e^{-f(x)} + f(x)\)

Hint:

To solve such differential equations, look for terms that approach zero as x increases. The limits often simplify the function form.

Answer 1

The Correct Option is B

1. Step 1: Begin with the given differential equation:

\[ y' + y f'(x) - f(x) f'(x) = 0 \]

1. This is a first-order linear differential equation involving both \( y \) and \( f(x) \).
2. Step 2: Reorganize the terms to isolate \( y' \):

\[ y' = f'(x) \left( f(x) - y \right) \]

1. The equation now describes how \( y \) changes with respect to \( x \), using the function \( f(x) \) and its derivative.
2. Step 3: Observe that the equation's structure implies a possible relationship between \( y \) and \( f(x) \). Given that \( f(x) \to 0 \) as \( x \to \infty \), we anticipate \( y(x) \) to simplify, potentially to a constant.
3. Step 4: Assume that as \( x \to \infty \), the functions' behavior becomes simpler. Since \( \lim_{x \to \infty} f(x) = 0 \), we hypothesize that \( y(x) \) might decay similarly to an exponential function.
4. Step 5: Propose a solution for \( y(x) \) in the form:

\[ y + 1 = e^{-f(x)} + f(x) \]

1. This form satisfies the differential equation and the condition that \( y(x) \to 0 \) as \( x \to \infty \) because \( f(x) \to 0 \).
2. Step 6: Substitute this proposed solution into the original differential equation to confirm it. If both sides of the equation are equal, our assumption is valid.

\[ y' + y f'(x) - f(x) f'(x) = 0 \]

1. Substituting \( y = e^{-f(x)} + f(x) - 1 \) into this equation validates the solution.
2. Step 7: Therefore, the correct relationship is:

\[ y + 1 = e^{-f(x)} + f(x) \]