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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:
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Verified Answer
The correct answer is:
$y(x)$ is decreasing in $\left(\frac{1}{2}, 1\right)$
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, $y(1)=\frac{1}{2} e^{-2}$
$\therefore \quad \frac{1}{2} e^{-2} \cdot e^{2} \cdot 1=\frac{1}{2}+c \Rightarrow c=0$
$\therefore \quad y(x)=\frac{x^{2}}{2} \cdot \frac{e^{-2 x}}{x}$
$y(x)=\frac{x}{2} \cdot e^{-2 x}$
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 $\left(\frac{1}{2}, 1\right)$
$\frac{d y}{d x}+\left(2+\frac{1}{x}\right) y=e^{-2 x}, x>0$
$\mathrm{IF}=e^{\int\left(2+\frac{1}{x}\right) d x}=e^{2 x+\ln x}=x e^{2 x}$
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, $y(1)=\frac{1}{2} e^{-2}$
$\therefore \quad \frac{1}{2} e^{-2} \cdot e^{2} \cdot 1=\frac{1}{2}+c \Rightarrow c=0$
$\therefore \quad y(x)=\frac{x^{2}}{2} \cdot \frac{e^{-2 x}}{x}$
$y(x)=\frac{x}{2} \cdot e^{-2 x}$
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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