Let $y=y(x)$ be the solution of the differential equation $(1-x^{2}) dy=[xy+(x^{3}+2) \sqrt{3(1-x^{2})}] dx$, $-1<x<1$, $y(0)=0$. If $y\left(\frac{1}{2}\right)=\frac{m}{n}$, $m$ and $n$ are coprime numbers, then $m+n$ is equal to ______.

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

(1 - x^{2}) d y = [x y + (x^{3} + 2) 3 (1 - x^{2})] d x

- 1 < x < 1

y (0) = 0

. If

y (\frac{1}{2}) = \frac{m}{n}

m

and

n

are coprime numbers, then

m + n

is equal to ______.

Answer: 97

\frac{d y}{d x} - \frac{x y}{1 - x ^{2}} = \frac{( x ^{3} + 2 ) 3}{1 - x ^{2}}

, integrating factor

= 1 - x^{2}

y 1 - x^{2} = 3 \int (x^{3} + 2) d x = 3 (\frac{x ^{4}}{4} + 2 x) + C

y (0) = 0 \Rightarrow C = 0

.

At

x = \frac{1}{2}

y (\frac{1}{2}) \frac{3}{4} = 3 (\frac{1}{64} + 1) = 3 \cdot \frac{65}{64}

, so

y (\frac{1}{2}) = \frac{65}{32}

m = 65, n = 32

(coprime),

m + n = 97

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Integer answer

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 →