If \(y=y(x)\) is the solution of the equation \(e^{\sin y} \cos y \frac{d y}{d x}+e^{\sin y} \cos x=\cos x, y(0)=0\); then \(1+y\left(\frac{\pi}{6}\right)+\frac{\sqrt{3}}{2} y\left(\frac{\pi}{3}\right)+\frac{1}{\sqrt{2}} y\left(\frac{\pi}{4}\right)\) is equal to ____

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Differential Equations

y = y (x)

is the solution of the equation

e^{s i n y} cos y \frac{d y}{d x} + e^{s i n y} cos x = cos x, y (0) = 0

; then

1 + y (\frac{π}{6}) + \frac{3}{2} y (\frac{π}{3}) + \frac{1}{2} y (\frac{π}{4})

is equal to ____

Answer: 1

Put

u = e^{s i n y}

, so that

\frac{d u}{d x} = e^{s i n y} cos y \frac{d y}{d x}

and the equation becomes linear in

u

\frac{d u}{d x} + u cos x = cos x

Its integrating factor is

e^{s i n x}

, and

\frac{d}{d x} (u e^{s i n x}) = cos x e^{s i n x} ⟹ u e^{s i n x} = e^{s i n x} + A

u = 1 + A e^{- s i n x}

. The condition

y (0) = 0

gives

u = e^{0} = 1

x = 0

, hence

A = 0

and

e^{s i n y} = 1

for every

x

.
Then

sin y = 0

, and a continuous solution with

y (0) = 0

y (x) = 0

throughout, so

1 + y (\frac{π}{6}) + \frac{3}{2} y (\frac{π}{3}) + \frac{1}{2} y (\frac{π}{4}) = 1 + 0 + 0 + 0 = 1

The value of the expression is

1

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Integer answer