JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{3}{\cos ^2 x} y=\frac{1}{\cos ^2 x}, x \in\left(\frac{-\pi}{3}, \frac{\pi}{3}\right)

, and

y\left(\frac{\pi}{4}\right)=\frac{4}{3}

, then

y\left(-\frac{\pi}{4}\right)

equals:

View answer and explanation

Correct option: A

\frac{1}{3}+e^6

Option A (

\frac{1}{3}+e^6

) is correct because it is the value of y(

-\frac{\pi}{4}

) obtained after solving the given differential equation and applying the initial condition y(

\frac{\pi}{4}

) =

\frac{4}{3}

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