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Differential Equations question

2020 · 4 Sep · Shift 2 · Q39

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JEE Main
Mathematics
Differential Equations
2020 · 4 Sep · Shift 2 · Q39

Differential Equations question

2020 · 4 Sep · Shift 2 · Q39

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

The solution of the differential equation

{{dy} \over {dx}} - {{y + 3x} \over {{{\log }_e}\left( {y + 3x} \right)}} + 3 = 0

is: (where c is a constant of integration)

A
$x - {1 \over 2}{\left( {{{\log }_e}\left( {y + 3x} \right)} \right)^2} = C$
B
$y + 3x - {1 \over 2}{\left( {{{\log }_e}x} \right)^2} = C$
C
x – loge(y+3x) = C
D
x – 2loge(y+3x) = C

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