JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

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Application +4 | -1

The solution curve of the differential equation

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

, which passes through th point

(0,1)

View answer and explanation

Correct option: D

y^2=1+y\left(\log _e\left(\frac{1+e^x}{2}\right)\right)

Option D (

y^2=1+y\left(\log _e\left(\frac{1+e^x}{2}\right)\right)

) is correct because it is the solution to the given differential equation that satisfies the initial condition (0,1).

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y=y(x)

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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)