Question

Let $y(x)$ be the solution of the differential equation $2 x^{2} d y+\left(e^{y}-2 x\right) d x=0, x>0 .$ If $y(e)=1$, then $y (1)$ is equal to:

• A. $0$
• B. $2$
• C. $\log _{ e } 2$
• D. $\log _{ e } (2e)$

Answer 1

Answer: (C)

$2 x^{2} d y+\left(e^{y}-2 x\right) d x=0$
$\frac{d y}{d x}+\frac{e^{y}-2 x}{2 x^{2}}=0 $
$\Rightarrow \frac{d y}{d x}+\frac{e^{y}}{2 x^{2}}-\frac{1}{x}=0$
$e^{-y} \frac{d y}{d x}-\frac{e^{-y}}{x}=-\frac{1}{2 x^{2}}$
$ \Rightarrow $ Put $e^{-y}=z$
$\frac{-d z}{d x}-\frac{z}{x}=-\frac{1}{2 x^{2}} $
$\Rightarrow x d z+z d x=\frac{d x}{2 x}$
$d(x z)=\frac{d x}{2 x}$
$ \Rightarrow x z=\frac{1}{2} \log _{e} x+c$
$x e^{-y}=\frac{1}{2} \log _{e} x+c$,
passes through $(e, 1)$
$\Rightarrow C =\frac{1}{2}$
$xe ^{- y }=\frac{\log _{ e } ex }{2}$
$e ^{- y }=\frac{1}{2} $
$\Rightarrow y =\log _{ e } 2$