Question

If the solution of the differential equation $\frac{dy}{dx}-y=1=-e^{-x}$ and $y\left(0\right)=y_0$ has a finite value, where $x\to \infty ,$ then
the value of $|\frac{2}{y_0}|$ is

Answer 1

$\frac{dy}{dx}-y=1-e^{-x}$
$\Rightarrow y{e}^{-x}=-e^{-x}+\frac{1}{2}{e}^{-2x}+c$
$y=-1+\frac{1}{2}{e}^{-x}+c{e}^x$
$x=0,y=y_0$
$so,c=y_0+\frac{1}{2}$
$y=-1+\frac{1}{2}{e}^{-x}+(y_0+\frac{1}{2})e^x$
$x\to \infty ,y\to$ finite value so $y_0+\frac{1}{2}=0$

$y_0=-\frac{1}{2}$

$\therefore |\frac{2}{y_0}|=4$