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 $y(1)= \frac{1}{2} \mathrm{e}^{-2}$, then:

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

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

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

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

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Differential Equations

y (x)

is the solution of the differential equation

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

where

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

, then:

Answer: (C)

The integrating factor is

e^{\int (2 + \frac{1}{x}) d x} = x e^{2 x}

. Multiplying through,

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

, so

y x e^{2 x} = \frac{x ^{2}}{2} + c

. The condition

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

gives

\frac{1}{2} e^{- 2} \cdot e^{2} = \frac{1}{2} + c

, so

c = 0

and

y = \frac{x}{2} e^{- 2 x}

. Then

\frac{d y}{d x} = \frac{e ^{- 2 x}}{2} (1 - 2 x)

, which is negative for

x > \frac{1}{2}

, so

y

is decreasing on

(\frac{1}{2}, 1)

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option