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Let $f$ be a differentiable function with $\lim _{x \rightarrow \infty} f(x)=0 .$ If $y^{\prime}+y f^{\prime}(x)-f(x) f^{\prime}(x)=0, \lim _{x \rightarrow \infty} y(x)=0$, then $\left(\right.$ where $\left.y^{\prime}=\frac{d y}{d x}\right)$
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Verified Answer
The correct answer is:
$y+1=e^{-f(x)}+f(x)$
Hint:
$\frac{d y}{d x}+f^{\prime}(x) y=f^{\prime}(x) f(x)$
$\Rightarrow y \times e^{f(x)}=\int f^{\prime}(x) f(x) e^{f(x)} d x$
$\Rightarrow y \times e^{f(x)}=e^{f(x)}(f(x)-1)+c \quad[$ Putting $f(x)=0 ; y=0, c=1]$
$\Rightarrow y \times e^{f(x)}=e^{f(x)}(f(x)-1)+1$
$\Rightarrow y=f(x)-1+e^{-f(x)}$
$\Rightarrow y+1=e^{-f(x)}+f(x)$
$\frac{d y}{d x}+f^{\prime}(x) y=f^{\prime}(x) f(x)$
$\Rightarrow y \times e^{f(x)}=\int f^{\prime}(x) f(x) e^{f(x)} d x$
$\Rightarrow y \times e^{f(x)}=e^{f(x)}(f(x)-1)+c \quad[$ Putting $f(x)=0 ; y=0, c=1]$
$\Rightarrow y \times e^{f(x)}=e^{f(x)}(f(x)-1)+1$
$\Rightarrow y=f(x)-1+e^{-f(x)}$
$\Rightarrow y+1=e^{-f(x)}+f(x)$
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