Question

If $y\left(x\right)$ is the solution of the differential equation $\frac{dy}{dx}+\left(\frac{2x+1}{x}\right)y=e^{-2x},x>0,$ where $y\left(1\right)=\frac{1}{2}{e}^{-2},$ then:

• A. $y\left({\log }_{e}2\right)=\frac{{\log }_{e}2}{4}$
• B. $y\left({\log }_{e}2\right)={\log }_{e}4$
• C. $y\left(x\right)$ is decreasing in $(0,1)$
• D. $y\left(x\right)$ is decreasing in $(\frac{1}{2},1)$

Answer 1

The correct option is D $y\left(x\right)$ is decreasing in $(\frac{1}{2},1)$

$\frac{dy}{dx}+\left(\frac{2x+1}{x}\right)y=e^{-2x},x>0$
$I.F.=e^{\int (\frac{2x}{x}+\frac{1}{x})dx}=x{e}^{2x}$

$\therefore y\times x{e}^{2x}=\int xdx$
$\Rightarrow y\times x{e}^{2x}=\frac{x^2}{2}+c$
Since $y\left(1\right)=\frac{1}{2}{e}^{-2},$
$\Rightarrow \frac{1}{2}{e}^{-2}\times e^2=\frac{1}{2}+c\Rightarrow c=0$
$\therefore y=\frac{x{e}^{-2x}}{2}$
$\Rightarrow y\left({\log }_{e}2\right)=\frac{\left({\log }_{e}2\right)e^{-2{\log }_{e}2}}{2}=\frac{{\log }_{e}2}{8}$
$\frac{dy}{dx}=\frac{1}{2}{e}^{-2x}(-2x+1)$
$\Rightarrow y\left(x\right)$ is decreasing in $(\frac{1}{2},1)$