Question

Let $f$ be a differentiable function with $\underset{x\to \infty }{lim}f\left(x\right)=0.$ If $y^{\prime }+y{f}^{\prime }\left(x\right)-f\left(x\right)f^{\prime }\left(x\right)=0,$ $\underset{x\to \infty }{lim}y\left(x\right)=0,$ then
$($where $y^{\prime }\equiv \frac{dy}{dx})$

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

Answer 1

The correct option is C $y+1=e^{-f\left(x\right)}+f\left(x\right)$

$\frac{dy}{dx}+y{f}^{\prime }\left(x\right)=f\left(x\right)f^{\prime }\left(x\right)I.F=e^{\int f^{\prime }\left(x\right)dx}=e^{f\left(x\right)}y.e^{f\left(x\right)}=\int e^{f\left(x\right)}f\left(x\right)f^{\prime }\left(x\right)dx$

Let $f\left(x\right)=t\Rightarrow f^{\prime }\left(x\right)dx=dt$
$\Rightarrow y{e}^{f\left(x\right)}=\int e^{t}tdt\Rightarrow y{e}^{f\left(x\right)}=e^t(t-1)+c\Rightarrow y{e}^{f\left(x\right)}=e^{f\left(x\right)}\left(f\right(x)-1)+c$

Given, $\underset{x\to \infty }{lim}f\left(x\right)=0,\underset{x\to \infty }{lim}y\left(x\right)=0$
$\Rightarrow 0.e^0=e^0(0-1)+c\Rightarrow c=1\therefore y{e}^{f\left(x\right)}=e^{f\left(x\right)}\left(f\right(x)-1)+1\Rightarrow y=f\left(x\right)-1+e^{-f\left(x\right)}$
$\Rightarrow y+1=f\left(x\right)+e^{-f\left(x\right)}$