JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

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Application +4 | -1

The solution of the differential equation

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

is (where

C

is a constant of integration)

View answer and explanation

Correct option: D

x - \frac{1}{2}{\left( {{{\log }_e}\,\left( {y + 3x} \right)} \right)^2} = C

The correct answer is D (

x - \frac{1}{2}({\log _e}(y + 3x))^2 = C

) because it is the solution to the given differential equation, obtained through substitution and integration.

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y=y(x)

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x = A\cos 4t + B\sin 4t,

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y=y(t)

\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

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\lim \limits_{{x \rightarrow \infty}} y(t)