Let $y=y(x)$ be the solution of the differential equation $\cos x \frac{d y}{d x}+2 y \sin x=\sin 2 x, x \in\left(0, \frac{\pi}{2}\right)$. If $y(\pi / 3)=0$, then $y(\pi / 4)$ is equal to:

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

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

.
If

y (π /3) = 0

, then

y (π /4)

is equal to:

Answer: (D)

Dividing through by

cos x

gives the linear form

\frac{d y}{d x} + 2 tan x y = 2 sin x

. The integrating factor is

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

, and multiplying through turns the left side into

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

. Integrating,

y sec^{2} x = \int 2 sin x sec^{2} x d x = 2 sec x + C

. The condition

y (π /3) = 0

gives

0 = 2 (2) + C

, so

C = - 4

and

y sec^{2} x = 2 sec x - 4

. At

x = \frac{π}{4}

, where

sec x = 2

2 y = 22 - 4

, so

y (\frac{π}{4}) = 2 - 2

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0$ , $y (0) = 1$ . Then the value of $y (1)$ is :
JEE Main 2026View solution →
Let $y = y (x)$ be the solution of the differential equation $sec x \frac{dy}{dx} - 2 y = 2 + 3 sin x, x \in (- \frac{π}{2}, \frac{π}{2})$ , $y (0) = - \frac{7}{4}$ . Then $y (\frac{π}{6})$ is equal to:
JEE Main 2026View solution →
Let the solution curve of the differential equation $x d y - y d x = x^{2} + y^{2} d x, x > 0, y (1) = 0$ , be $y = y (x)$ . Then $y (3)$ is equal to
JEE Main 2026View solution →
If $y = y (x)$ satisfies the differential equation
$16 (x + 9 x) (4 + 9 + x) cos y d y = (1 + 2 sin y) d x$ , $x > 0$ and $y (256) = \frac{π}{2}$ , $y (49) = α$ , then $2 sin α$ is equal to :
JEE Main 2026View solution →
Let $y = y (x)$ be the solution of the differential equation
$x^{4} d y + (4 x^{3} y + 2 sin x) d x = 0$ , $x > 0$ , $y (\frac{π}{2}) = 0$ .

Then $π^{4} y (\frac{π}{3})$ is equal to :
JEE Main 2026View solution →
If the solution curve $y = f (x)$ of the differential equation $(x^{2} - 4) y^{'} - 2 x y + 2 x (4 - x^{2})^{2} = 0, x > 2$ , passes through the point (3,15), then the local maximum value of f is
JEE Main 2026View solution →