Question

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

Answer 1

Answer: (B)

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

$\frac{dy}{dx}+\left(\frac{2x+1}{x}\right)y=e^{-2x}$
I.F. $=e^{\int \left(\frac{2x+1}{x}\right)dx}=e^{\int (2+\frac{1}{x})dx}=e^{2x+\ell nx}=e^{2x}.x$
So, $y\left(x{e}^{2x}\right)=\int e^{-2x}.x{e}^{2x}+C$
$⟹xy{w}^{2x}=\int xdx+C$
$⟹2xy{e}^{2x}=x^2+2C$
It passes through $(1,\frac{1}{2}{e}^{-2})$ we get $C=0$
$y=\frac{x{e}^{-2x}}{2}$
$⟹\frac{dy}{dx}=\frac{1}{2}{e}^{-2x}(-2x+1)$
$⟶f\left(x\right)$ is decreasing in $(\frac{1}{2},1)$
$y\left(lo{g}_{e}2\right)=\frac{\left(lo{g}_{e}2\right)e^{-2\left(lo{g}_{e}2\right)}}{2}$
$=\frac{1}{8}lo{g}_{e}2$