Question

Let $y=g\left(x\right)$ be the solution of differential equation $\frac{dy}{dx}+y=\frac{2x{e}^{-x}}{1+y{e}^x}$ such that $g\left(0\right)=1.$ If $[.]$ denotes the greatest integer function, then $\left[\frac{g(-1)}{e}\right]$ is equal to

Answer 1

$y^{\prime }+y=\frac{2x{e}^{-x}}{1+y{e}^x}\Rightarrow \frac{dy}{dx}+y=\frac{1}{e^x}\cdot \frac{2x}{1+y{e}^x}\Rightarrow e^x\frac{dy}{dx}+e^{x}y=\frac{2x}{1+y{e}^x}$
Let $e^{x}y=t$
$\Rightarrow e^x\frac{dy}{dx}+y{e}^x=\frac{dt}{dx}$
Then $\frac{dt}{dx}=\frac{2x}{1+t}$
$\Rightarrow \int (1+t)dt=\int 2xdx\Rightarrow t+\frac{t^2}{2}=x^2+c\Rightarrow e^{x}y+\frac{(e^{x}y{)}^2}{2}=x^2+c$
Since $g\left(0\right)=1,$
$\Rightarrow c=\frac{3}{2}\therefore e^{x}y+\frac{(e^{x}y{)}^2}{2}=x^2+\frac{3}{2}\Rightarrow \frac{(1+y{e}^x{)}^2}{2}=x^2+2\Rightarrow y{e}^x+1=\pm \sqrt{2x^2+4}$
$y\left(0\right)=1$ is possible when we take positive sign.
$\Rightarrow y{e}^x+1=\sqrt{2x^2+4}\therefore g(-1)=(\sqrt{6}-1)e$
Hence, $\left[\frac{g(-1)}{e}\right]=1$