If $\frac{d y}{d x}+\frac{3}{\cos ^{2} x} y=\frac{1}{\cos ^{2} x}, x \in\left(\frac{-\pi}{3}, \frac{\pi}{3}\right)$ and $y\left(\frac{\pi}{4}\right)=\frac{4}{3}$, then $y\left(-\frac{\pi}{4}\right)$ equals:

(A) $\frac{1}{3}+\mathrm{e}^{6}$

(D) $\frac{1}{3}+\mathrm{e}^{3}$

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Differential Equations

\frac{d y}{d x} + \frac{3}{c o s ^{2} x} y = \frac{1}{c o s ^{2} x}, x \in (\frac{- π}{3}, \frac{π}{3})

and

y (\frac{π}{4}) = \frac{4}{3}

, then

y (- \frac{π}{4})

equals:

Answer: (A)

This is a linear first-order equation,

\frac{d y}{d x} + 3 sec^{2} x y = sec^{2} x

, with integrating factor

e^{\int 3 s e c^{2} x d x} = e^{3 t a n x}

. Multiplying through makes the left side

\frac{d}{d x} (y e^{3 t a n x}) = sec^{2} x e^{3 t a n x}

, and integrating gives

y e^{3 t a n x} = \frac{1}{3} e^{3 t a n x} + C

, so

y = \frac{1}{3} + C e^{- 3 t a n x}

. The condition

y (\frac{π}{4}) = \frac{4}{3}

with

tan \frac{π}{4} = 1

fixes

\frac{4}{3} = \frac{1}{3} + C e^{- 3}

, so

C = e^{3}

. At

x = - \frac{π}{4}

, where

tan = - 1

y = \frac{1}{3} + e^{3} \cdot e^{3} = \frac{1}{3} + e^{6}

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option