Question

Let $y(x)$ be the solution of the differential equation $2x^{2}dy+\left(e^y-2x\right)dx=0,x>0.$ 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)

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