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: Start 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: Rearrange the terms to isolate \( y' \):

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

1. Now, we have an equation that describes how \( y \) changes with respect to \( x \), in terms of the function \( f(x) \) and its derivative.
2. Step 3: Notice that the form of the equation suggests that \( y \) might be related to \( f(x) \) in some way. Since \( f(x) \to 0 \) as \( x \to \infty \), we expect \( y(x) \) to simplify as well, possibly to a constant.
3. Step 4: Assume that as \( x \to \infty \), the behavior of the functions becomes simpler. Specifically, since \( \lim_{x \to \infty} f(x) = 0 \), we hypothesize that \( y(x) \) may behave similarly to an exponential function that decays as \( f(x) \) does.
4. Step 5: Assume 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 as \( x \to \infty \), \( y(x) \to 0 \) because \( f(x) \to 0 \).
2. Step 6: Substitute this proposed solution into the original differential equation to verify that it satisfies the equation. If both sides of the equation balance, then our assumption is correct.

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

1. After substituting \( y = e^{-f(x)} + f(x) - 1 \) into this equation, we find that it holds true, confirming the correctness of our solution.
2. Step 7: Therefore, the correct relationship is:

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

Answer 2

Solving a First-Order Linear Differential Equation

We are given a differential function $f(x)$ with $\lim_{x \to \infty} f(x) = 0$. The function $y$ satisfies the differential equation $y' + yf'(x) - f(x)f'(x) = 0$, and $\lim_{x \to \infty} y(x) = 0$. We need to find the relationship between $y$ and $f(x)$.

Step 1: Rewrite the differential equation

The given differential equation is:

$$ y' + yf'(x) - f(x)f'(x) = 0 $$

Rewrite it in the standard form of a first-order linear differential equation $\frac{dy}{dx} + P(x)y = Q(x)$:

$$ \frac{dy}{dx} + f'(x)y = f(x)f'(x) $$

Here, $P(x) = f'(x)$ and $Q(x) = f(x)f'(x)$.

Step 2: Find the integrating factor (IF)

The integrating factor is given by $IF = e^{\int P(x) dx}$:

$$ IF = e^{\int f'(x) dx} = e^{f(x)} $$

Step 3: Multiply the differential equation by the integrating factor

$$ e^{f(x)} \frac{dy}{dx} + e^{f(x)} f'(x)y = e^{f(x)} f(x)f'(x) $$

Step 4: Recognize the left side as the derivative of a product

The left side is the derivative of $ye^{f(x)}$ with respect to $x$:

$$ \frac{d}{dx}(ye^{f(x)}) = e^{f(x)} \frac{dy}{dx} + y e^{f(x)} f'(x) $$

Step 5: Integrate both sides with respect to $x$

$$ \int \frac{d}{dx}(ye^{f(x)}) dx = \int e^{f(x)} f(x)f'(x) dx $$

$$ ye^{f(x)} = \int e^{f(x)} f(x) d(f(x)) $$

Let $u = f(x)$, then $du = d(f(x))$. The integral becomes $\int ue^u du$. Using integration by parts, $\int ue^u du = ue^u - e^u + C$.

Substituting back $f(x)$:

$$ ye^{f(x)} = f(x)e^{f(x)} - e^{f(x)} + C $$

Step 6: Solve for $y$

Divide by $e^{f(x)}$:

$$ y = f(x) - 1 + Ce^{-f(x)} $$

$$ y + 1 = f(x) + Ce^{-f(x)} $$

Step 7: Use the limit condition $\lim_{x \to \infty} y(x) = 0$

$$ \lim_{x \to \infty} (y(x) + 1) = \lim_{x \to \infty} (f(x) + Ce^{-f(x)}) $$

Given $\lim_{x \to \infty} f(x) = 0$ and $\lim_{x \to \infty} y(x) = 0$:

$$ 0 + 1 = 0 + C e^{-0} $$

$$ 1 = C \cdot 1 \implies C = 1 $$

Step 8: Substitute the value of C back into the equation

$$ y + 1 = f(x) + 1 \cdot e^{-f(x)} $$

$$ y + 1 = f(x) + e^{-f(x)} $$

Final Answer: (B) $y + 1 = e^{-f(x)} + f(x)$