JEE Math Practice Question
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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