Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}+\frac{\sqrt{2} y}{2 \cos ^{4} x-\cos 2 x}=\mathrm{xe}^{\tan ^{-1}(\sqrt{2} \cot 2 \mathrm{x})}, 0<x<\pi / 2$ with $y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{32}$. If $y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{18} e^{-\tan ^{-1}(\alpha)}$, then the value of $3 \alpha^{2}$ is equal to ____.

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

\frac{d y}{d x} + \frac{2 y}{2 c o s ^{4} x - c o s 2 x} = xe^{t a n^{- 1} (2 c o t 2 x)}, 0 < x < π /2

with

y (\frac{π}{4}) = \frac{π ^{2}}{32}

.
If

y (\frac{π}{3}) = \frac{π ^{2}}{18} e^{- t a n^{- 1} (α)}

, then the value of

3 α^{2}

is equal to ____.

Answer: 2

The integrating factor is

I F = e^{- t a n^{- 1} (2 c o t 2 x)}

.

Multiplying the DE by IF and integrating:

y e^{- t a n^{- 1} (2 c o t 2 x)} = \int x d x = \frac{x ^{2}}{2} + c

Using

y (\frac{π}{4}) = \frac{π ^{2}}{32}

and noting that

tan^{- 1} (2 cot (π /2)) = 0

\frac{π ^{2}}{32} = \frac{π ^{2}}{32} + c \Rightarrow c = 0

So:

y = \frac{x ^{2}}{2} e^{t a n^{- 1} (2 c o t 2 x)}

.

At

x = \frac{π}{3}

y (\frac{π}{3}) = \frac{π ^{2}}{18} e^{t a n^{- 1} (2 c o t (2 π /3))} = \frac{π ^{2}}{18} e^{t a n^{- 1} (- 2 / 3)} = \frac{π ^{2}}{18} e^{- t a n^{- 1} (2/3)}

Comparing:

α = \frac{2}{3}

.

Therefore:

3 α^{2} = 3 \cdot \frac{2}{3} = 2

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Integer answer