Question

The solution of $x\frac{dx}{dy}=y+xe^{y/x}$ with $y(1) = 0$ is

• A. $e^{y/x}+\log\,x=1$
• B. $e^{-y/x}=\log\,x$
• C. $e^{-y/x}+2\log\,x=1$
• D. $e^{-y/x}+\log\,x=1$

Answer 1

Answer: (A)

$x \frac{dy}{dx} = y + xe^{yx}$
$\Rightarrow \frac{dy}{dx}- \frac{y}{x} = e^{yx}$
Put $y = vx$
$\Rightarrow \frac{dy}{dx} = \upsilon+x \frac{d\upsilon}{dx}$
$\therefore v + x \frac{dv}{dx}-\upsilon = e^{\upsilon}$
$\Rightarrow x \frac{dv}{dx} = e^{\upsilon}$
$\Rightarrow \int e^{-\upsilon} = \int \frac{dx}{x}+c$
$\Rightarrow -e^{-\upsilon} = log\,x+c$
$\Rightarrow e^{-yx} = log\, x + c$
When $x = 1$, $y = 0$
$\therefore -e^{-0} = log 1 + c \Rightarrow -1 = 0+c \Rightarrow c = -1$
$\therefore x = 1$, $y = 0$
$\therefore -e^{-yx} = log\,x-1$
$\Rightarrow log\, x + e^{-yx} = 1$