JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Differential Equations

Let

y = y (x)

be in the solution of the differential equation

cos x (3 sin x + cos x + 3) d y = (1 + y sin x (3 sin x + cos x + 3)) d x, 0 \leq x \leq \frac{π}{2}, y (0) = 0

. Then

y (\frac{π}{3})

is equal to:

Answer: (A)

Solution

Dividing throughout by

cos x (3 sin x + cos x + 3)

puts the equation in linear form:

\frac{d y}{d x} - y tan x = \frac{1}{cos x ( 3 sin x + cos x + 3 )}

The integrating factor is

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

, and multiplying by it gives

\frac{d}{d x} (y cos x) = \frac{1}{3 sin x + cos x + 3}

With

t = tan \frac{x}{2}

, the denominator is

\frac{2 ( t + 1 ) ( t + 2 )}{1 + t ^{2}}

and

d x = \frac{2 d t}{1 + t ^{2}}

, so

y cos x = \int \frac{d t}{( t + 1 ) ( t + 2 )} = lo g_{e} \frac{t + 1}{t + 2} + C

x = 0

both

y

and

t

vanish, so

C = lo g_{e} 2

. At

x = \frac{π}{3}

t = \frac{1}{3}

and

cos x = \frac{1}{2}

, giving

\frac{y}{2} = lo g_{e} \frac{2 ( 1 + 3 )}{1 + 2 3} = lo g_{e} \frac{2 3 + 10}{11}

Hence

y (\frac{π}{3}) = 2 lo g_{e} (\frac{2 3 + 10}{11})

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option