If y(x) is the solution of the differential e | vedclass

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(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = 2 + \frac{1}{x}$ and $Q

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

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(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = 2 + \frac{1}{x}$ and $Q(x) = e^{-2x}$.The Integrating Factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int (2 + \frac{1}{x}) dx} = e^{2x + \log_e x} = x e^{2x}$.The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.$y(x e^{2x}) = \int e^{-2x} \cdot (x e^{2x}) dx + C = \int x dx + C = \frac{x^2}{2} + C$.Given $y(1) = \frac{1}{2}e^{-2}$,we substitute $x=1$ and $y=\frac{1}{2}e^{-2}$:$\frac{1}{2}e^{-2} \cdot (1 \cdot e^2) = \frac{1^2}{2} + C \Rightarrow \frac{1}{2} = \frac{1}{2} + C \Rightarrow C = 0$.Thus,$y = \frac{x}{2}e^{-2x}$.To check the monotonicity,we find $\frac{dy}{dx} = \frac{1}{2} e^{-2x} + \frac{x}{2} (-2 e^{-2x}) = \frac{e^{-2x}}{2} (1 - 2x)$.For $x \in \left( \frac{1}{2}, 1 \right)$,$1 - 2x < 0$,so $\frac{dy}{dx} < 0$,which means $y(x)$ is decreasing in $\left( \frac{1}{2}, 1 \right)$.

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

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