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Differential Equations question

2019 · 11 Jan · Shift 1 · Q24
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  5. /2019 · 11 Jan · Shift 1 · Q24

Differential Equations question

2019 · 11 Jan · Shift 1 · Q24

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If y(x) is the solution of the differential equation dydx+(2x+1x)y=e−2x,  x>0, {{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}},\,\,x \gt 0,\,dxdy​+(x2x+1​)y=e−2x,x>0, where y(1)=12e−2,y\left( 1 \right) = {1 \over 2}{e^{ - 2}},y(1)=21​e−2, then
  1. A
    y(loge2) = loge4
  2. B
    y(x) is decreasing in (0, 1)
  3. C
    y(loge2) = log⁡e24{{{{\log }_e}2} \over 4}4loge​2​
  4. D
    y(x) is decreasing in (12,1)\left( {{1 \over 2},1} \right)(21​,1)
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