Let \(y=y(x)\) be solution of the following differential equation \(e^{y} \frac{d y}{d x}-2 e^{y} \sin x+\sin x \cos ^{2} x=0, y\left(\frac{\pi}{2}\right)=0\). If \(y(0)=\log _{e}\left(\alpha+\beta e^{-2}\right)\), then \(4(\alpha+\beta)\) is equal to ____

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Differential Equations

Let

y = y (x)

be solution of the following differential equation

e^{y} \frac{d y}{d x} - 2 e^{y} sin x + sin x cos^{2} x = 0, y (\frac{π}{2}) = 0

.
If

y (0) = lo g_{e} (α + β e^{- 2})

, then

4 (α + β)

is equal to ____

Answer: 4

With

u = e^{y}

\frac{d u}{d x} = e^{y} \frac{d y}{d x}

, so the equation becomes linear in

u

\frac{d u}{d x} - 2 u sin x = - sin x cos^{2} x

The integrating factor is

e^{- \int 2 s i n x d x} = e^{2 c o s x}

, giving

\frac{d}{d x} (u e^{2 c o s x}) = - sin x cos^{2} x e^{2 c o s x}

With

t = cos x

the right side integrates as

\int t^{2} e^{2 t} d t = e^{2 t} (\frac{t ^{2}}{2} - \frac{t}{2} + \frac{1}{4})

, so

u e^{2 c o s x} = e^{2 c o s x} (\frac{cos ^{2} x}{2} - \frac{cos x}{2} + \frac{1}{4}) + C

x = \frac{π}{2}

cos x = 0

and

u = e^{0} = 1

, so

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

and

C = \frac{3}{4}

. At

x = 0

u e^{2} = e^{2} (\frac{1}{2} - \frac{1}{2} + \frac{1}{4}) + \frac{3}{4} \Rightarrow e^{y (0)} = \frac{1}{4} + \frac{3}{4} e^{- 2}

α = \frac{1}{4}

β = \frac{3}{4}

, and

4 (α + β) = 4

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Integer answer