If $y=y(x)$ is the solution of the differential equation, $\frac{d y}{d x}+2 y \tan x=\sin x, y\left(\frac{\pi}{3}\right)=0$, then the maximum value of the function $y(x)$ over $R$ is equal to :

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Differential Equations

y = y (x)

is the solution of the differential equation,

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

, then the maximum value of the function

y (x)

over

R

is equal to :

Answer: (B)

The equation is linear in

y

, with integrating factor

e^{\int 2 t a n x d x} = sec^{2} x

, so

\frac{d}{d x} (y sec^{2} x) = sin x sec^{2} x = sec x tan x

Integrating,

y sec^{2} x = sec x + C

, that is

y = cos x + C cos^{2} x

. From

y (\frac{π}{3}) = 0

\frac{1}{2} + \frac{C}{4} = 0

, so

C = - 2

and

y = cos x - 2 cos^{2} x

Putting

u = cos x

y = - 2 (u - \frac{1}{4})^{2} + \frac{1}{8}

and

u = \frac{1}{4}

does occur, since

\frac{1}{4}

lies in

[- 1, 1]

. The greatest value of

y

over

R

\frac{1}{8}

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option

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