If y(x) is the solution of the differential equation where y(1) = e

Engineering

Mathematics

Monotonicity

Question

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

y (log_e 2) = log_e4

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

$y ({log}_{e} 2) = \frac{\log_{e} 2}{4}$

y (x) is decreasing in (0, 1)

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$I . F . = e^{\int (2 + \frac{1}{x}) dx} = {xe}^{2x}$

solution is given by

${y xe}^{2x} = \int x dx + c \Rightarrow {yxe}^{2x} = \frac{x^{2}}{2} + c$

$y (1) = \frac{1}{2} e^{- 2} \Rightarrow c = 0$

$y = \frac{{xe}^{- 2 x}}{2}$

$y'^{} = \frac{e^{- 2 x}}{2} (1 - 2x) < 0 \forall x > \frac{1}{2}$