If for the solution curve $y=f(x)$ of the differential equation $\frac{d y}{d x}+(\tan x) y=\frac{2+\sec x}{(1+2 \sec x)^{2}}$, $x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right), f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{10}$, then $f\left(\frac{\pi}{4}\right)$ is equal to:

(A) $\frac{9 \sqrt{3}+3}{10(4+\sqrt{3})}$

(B) $\frac{\sqrt{3}+1}{10(4+\sqrt{3})}$

(C) $\frac{5-\sqrt{3}}{2 \sqrt{2}}$

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Differential Equations

If for the solution curve

y = f (x)

of the differential equation

\frac{d y}{d x} + (tan x) y = \frac{2 + s e c x}{( 1 + 2 s e c x ) ^{2}}

x \in (\frac{- π}{2}, \frac{π}{2}), f (\frac{π}{3}) = \frac{3}{10}

, then

f (\frac{π}{4})

is equal to:

Answer: (D)

The integrating factor is

e^{\int t a n x d x} = sec x

, so

y sec x = \int sec x \cdot \frac{2 + sec x}{( 1 + 2 sec x ) ^{2}} d x = \int \frac{2 cos x + 1}{( cos x + 2 ) ^{2}} d x

Using

t = tan \frac{x}{2}

(so

cos x = \frac{1 - t ^{2}}{1 + t ^{2}}

) this reduces, after simplification, to

\int \frac{2 ( 3 - t ^{2} )}{( t ^{2} + 3 ) ^{2}} d t

and substituting

u = t + \frac{3}{t}

gives

y sec x = \frac{2}{t + \frac{3}{t}} + C

x = \frac{π}{3}

t = tan \frac{π}{6} = \frac{1}{3}

, and

f (\frac{π}{3}) = \frac{3}{10}

gives

2 \cdot \frac{3}{10} = \frac{2}{\frac{1}{3} + 3 3} + C = \frac{2 3}{10} + C \Rightarrow C = 0

x = \frac{π}{4}

t = tan \frac{π}{8} = 2 - 1

y 2 = \frac{2}{( 2 - 1 ) + \frac{3}{2 - 1}} = \frac{2 ( 2 - 1 )}{( 2 - 1 ) ^{2} + 3} = \frac{2 ( 2 - 1 )}{6 - 2 2}

y = \frac{2 ( 2 - 1 )}{6 - 2 2} = \frac{2 - 2}{2 ( 3 - 2 )} = \frac{4 - 2}{14}

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option