COMPASSLearning

Talk to us Register Login

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Differential Equations

Let

\int_{0}^{x} 1 - (y^{'} (t))^{2} dt = \int_{0}^{x} y (t) dt, 0 \leq x \leq 3, y \geq 0

,

y (0) = 0

. Then at

x = 2, y^{''} + y + 1

is equal to :

(A)1✓ Correct
(B)2
(C) $2$
(D) $1/2$

Answer: (A)

Solution

Differentiating

\int_{0}^{x} 1 - (y^{'} (t))^{2} d t = \int_{0}^{x} y (t) d t

with respect to

x

:

1 - (y^{'})^{2} = y

So

(y^{'})^{2} = 1 - y^{2}

, i.e.

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

(taking

y

increasing from

y (0) = 0

,

y \geq 0

).

Integrating:

sin^{- 1} y = x + c

; with

y (0) = 0

,

c = 0

, so

y = sin x

.

Then

y^{''} = - sin x = - y

, so at

x = 2

:

y^{''} + y + 1 = - y + y + 1 = 1

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Differential Equations →

More JEE Main questions on Differential Equations

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 →