JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Differential Equations

y = y (x), y \in [0, \frac{π}{2}]

is the solution of the differential equation

sec y \frac{d y}{d x} - sin (x + y) - sin (x - y) = 0

, with

y (0) = 0

, then

5 y^{'} (\frac{π}{2})

is equal to ____ .

Answer: 2

Solution

Using

sin (x + y) + sin (x - y) = 2 sin x cos y

, the equation becomes

sec y \frac{d y}{d x} = 2 sin x cos y \Rightarrow sec^{2} y d y = 2 sin x d x

Integrating,

tan y = - 2 cos x + C

, and

y (0) = 0

gives

C = 2

, so

tan y = 2 (1 - cos x)

.
Differentiating this,

sec^{2} y y^{'} = 2 sin x

, so

y^{'} = 2 sin x cos^{2} y

.
At

x = \frac{π}{2}

tan y = 2

, hence

sec^{2} y = 5

and

cos^{2} y = \frac{1}{5}

, giving

y^{'} (\frac{π}{2}) = \frac{2}{5}

and

5 y^{'} (\frac{π}{2}) = 2

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