If $y=f(x)$ is the solution of the differential equation $\frac{d y}{d x}=(\tan x-y) \sec ^{2} x$, $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, such that $y(0)=0$, then $y\left(-\frac{\pi}{4}\right)$ is equal to

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Differential Equations

y = f (x)

is the solution of the differential equation

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

x \in (- \frac{π}{2}, \frac{π}{2})

, such that

y (0) = 0

, then

y (- \frac{π}{4})

is equal to

Answer: (C)

Write the equation as

\frac{d y}{d x} + y sec^{2} x = tan x sec^{2} x

and substitute

t = tan x

\frac{d t}{d x} = sec^{2} x

, to get

\frac{d y}{d t} + y = t

. With integrating factor

e^{t}

\frac{d}{d t} (y e^{t}) = t e^{t}

, so

y e^{t} = e^{t} (t - 1) + C

and

y = tan x - 1 + C e^{- t a n x}

. The condition

y (0) = 0

gives

C = 1

. At

x = - \frac{π}{4}

tan x = - 1

, so

y = - 1 - 1 + e = e - 2

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · 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 →