If y(x) is the solution of the differential equation
Question:

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

  1. (1) $y\left(\log _{e} 2\right)=\log _{e} 4$

  2. (2) $y\left(\log _{e} 2\right)=\frac{\log _{e} 2}{4}$

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

  4. (4) $\mathrm{y}(x)$ is decreasing in $(0,1)$


Correct Option: , 3

Solution:

Given differential equation is,

$\frac{d y}{d x}+\left(2+\frac{1}{x}\right) y=e^{-2 x}, x>0$

$\mathrm{IF}=e^{\int\left(2+\frac{1}{x}\right) d x}=e^{2 x+\ln x}=x e^{2 x}$

Complete solution is given by

$y(x) \cdot x e^{2 x}=\int x e^{2 x} \cdot e^{-2 x} d x+c$

$=\int x d x+c$

$y(x) \cdot e^{2 x} \cdot x=\frac{x^{2}}{2}+c$

Given, $y(1)=\frac{1}{2} e^{-2}$

$\therefore \quad \frac{1}{2} e^{-2} \cdot e^{2} \cdot 1=\frac{1}{2}+c \Rightarrow c=0$

$\therefore y(x)=\frac{x^{2}}{2} \cdot \frac{e^{-2 x}}{x}$

$y(x)=\frac{x}{2} \cdot e^{-2 x}$

Differentiate both sides with respect to $x$,

$y^{\prime}(x)=\frac{e^{-2 x}}{2}(1-2 x)<0 \forall x \in\left(\frac{1}{2}, 1\right)$\

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

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