JEE Main · 2016 · 03 Apr 2016 · Offline · Differential Equations

If a curve

y = f (x)

passes through the point (1,-1) and satisfies the differential equation,

y (1 + x y) d x = x d y

, then

f (- \frac{1}{2})

is equal to:

Answer: (C)

Solution

Group the equation

y (1 + x y) d x = x d y

\frac{x d y - y d x}{y ^{2}} = x d x

, and note the left side is

- d (\frac{x}{y})

. Integrating,

- \frac{x}{y} = \frac{x ^{2}}{2} + c

. The curve passes through

(1, - 1)

- \frac{1}{- 1} = \frac{1}{2} + c

, so

c = \frac{1}{2}

. Thus

- \frac{x}{y} = \frac{x ^{2} + 1}{2}

, giving

y = \frac{- 2 x}{x ^{2} + 1}

. At

x = - \frac{1}{2}

y = \frac{- 2 ( - 1/2 )}{1/4 + 1} = \frac{1}{5/4} = \frac{4}{5}

JEE Main · 2016 · 03 Apr 2016 · Offline · 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 →