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(B) The given differential equation is $\frac{dy}{dx} + y f'(x) = f(x) f'(x)$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)

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

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(B) The given differential equation is $\frac{dy}{dx} + y f'(x) = f(x) f'(x)$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = f'(x)$ and $Q(x) = f(x) f'(x)$.The integrating factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int f'(x) dx} = e^{f(x)}$.The solution is given by $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.$y e^{f(x)} = \int f(x) f'(x) e^{f(x)} dx + C$.Let $t = f(x)$, then $dt = f'(x) dx$.$y e^{f(x)} = \int t e^t dt + C = e^t(t-1) + C$.Substituting back $t = f(x)$, we get $y e^{f(x)} = e^{f(x)}(f(x)-1) + C$.Given $\lim_{x \rightarrow \infty} f(x) = 0$ and $\lim_{x \rightarrow \infty} y(x) = 0$, we substitute these limits into the equation:$0 \cdot e^0 = e^0(0-1) + C \Rightarrow 0 = -1 + C \Rightarrow C = 1$.Thus, $y e^{f(x)} = e^{f(x)}(f(x)-1) + 1$.Dividing by $e^{f(x)}$, we get $y = f(x) - 1 + e^{-f(x)}$.Rearranging gives $y + 1 = f(x) + e^{-f(x)}$.

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:

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