If $y(x)$ satisfies the differential equation $y^{\prime}-y \tan x=2 x \sec x$ and $y(0)=0$, then

(A) $y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8 \sqrt{2}}$; (D) $y^{\prime}\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}}$

(B) $y^{\prime}\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{18}$

(C) $y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{9}$

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Differential Equations

y (x)

satisfies the differential equation

y^{'} - y tan x = 2 x sec x

and

y (0) = 0

, then

Answer: (A), (D)

The differential equation is

\frac{d y}{d x} - y tan x = 2 x sec x

. Multiplying by

cos x

cos x \frac{d y}{d x} - y sin x = 2 x

, which gives

\frac{d}{d x} (y cos x) = 2 x

. Integrating:

y cos x = x^{2} + c

. From

y (0) = 0

, we get

c = 0

, so

y = \frac{x ^{2}}{c o s x}

.

(A)

y (\frac{π}{4}) = \frac{( π /4 ) ^{2}}{1/ 2} = \frac{π ^{2}}{8 2}

✓

(C)

y (\frac{π}{3}) = \frac{( π /3 ) ^{2}}{1/2} = \frac{2 π ^{2}}{9}

✗

y^{'} = \frac{2 x c o s x + x ^{2} s i n x}{c o s ^{2} x}

(B)

y^{'} (\frac{π}{4}) = \frac{π / 2 + π ^{2} / ( 16 2 )}{1/2} = \frac{π ^{2}}{8 2} + \frac{π}{2}

✗

(D)

y^{'} (\frac{π}{3}) = \frac{2 π /3 \cdot 1/2 + π ^{2} /9 \cdot 3 /2}{1/4} = \frac{4 π}{3} + \frac{2 π ^{2}}{3 3}

✓

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · One or more correct options