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(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = f^{\prime}(x)$ and $Q(

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

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(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = f^{\prime}(x)$ and $Q(x) = f(x)f^{\prime}(x)$.The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int f^{\prime}(x) dx} = e^{f(x)}$.The general solution is given by $y \cdot IF = \int Q(x) \cdot IF dx + C$.$y \cdot e^{f(x)} = \int f(x) f^{\prime}(x) e^{f(x)} dx + C$.Let $u = f(x)$, then $du = f^{\prime}(x) dx$. The integral becomes $\int u e^u du = u e^u - e^u$.Thus, $y \cdot e^{f(x)} = e^{f(x)}(f(x) - 1) + C$.Given $\lim_{x \rightarrow \infty} f(x) = 0$ and $\lim_{x \rightarrow \infty} y(x) = 0$, we substitute these into the equation:$0 \cdot e^0 = e^0(0 - 1) + C \Rightarrow 0 = -1 + C \Rightarrow C = 1$.Substituting $C = 1$ back into the equation:$y \cdot e^{f(x)} = e^{f(x)}(f(x) - 1) + 1$.Dividing by $e^{f(x)}$:$y = f(x) - 1 + e^{-f(x)}$.Rearranging gives $y + 1 = e^{-f(x)} + f(x)$.

Let $f$ be a differentiable function with $\lim_{x \rightarrow \infty} f(x) = 0$. If $y^{\prime} + y f^{\prime}(x) - f(x) f^{\prime}(x) = 0$ and $\lim_{x \rightarrow \infty} y(x) = 0$, then (where $y^{\prime} = \frac{dy}{dx}$):

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