If y(x) solves dydx+2x+1xy=e^-2x, x0, y(1)=12e^-2, then:

Question

If $y(x)$ solves $\dfrac{dy}{dx}+\dfrac{2x+1}{x}y=e^{-2x}$, $x>0$, $y(1)=\dfrac12e^{-2}$, then:

Accepted Answer

I.F.$=xe^{2x}$. $y(x)xe^{2x} = \dfrac{x^2}{2}+c$; using $y(1)$ gives $c=0$, so $y(x)=\dfrac{x}{2}e^{-2x}$.

$y'(x)=\dfrac{e^{-2x}}{2}(1-2x)<0$ for $x\in(1/2,1)$, so $y$ is decreasing there.

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