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JEE Advanced 2025 Paper 2 Online
Numerical
+4
-0

Let $y(x)$ be the solution of the differential equation

$$ x^2 \frac{d y}{d x}+x y=x^2+y^2, \quad x>\frac{1}{e} $$

satisfying $y(1)=0$. Then the value of $2 \frac{(y(e))^2}{y\left(e^2\right)}$ is ____________.

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2
JEE Advanced 2025 Paper 1 Online
Numerical
+4
-0

For all x > 0, let y₁(x), y₂(x), and y₃(x) be the functions satisfying

$ \frac{dy_1}{dx} - (\sin x)^2 y_1 = 0, \quad y_1(1) = 5, $

$ \frac{dy_2}{dx} - (\cos x)^2 y_2 = 0, \quad y_2(1) = \frac{1}{3}, $

$ \frac{dy_3}{dx} - \frac{(2-x^3)}{x^3} y_3 = 0, \quad y_3(1) = \frac{3}{5e}, $

respectively. Then

$ \lim\limits_{x \to 0^+} \frac{y_1(x)y_2(x)y_3(x) + 2x}{e^{3x} \sin x} $

is equal to __________________.

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3
JEE Advanced 2023 Paper 2 Online
Numerical
+4
-0
For $x \in \mathbb{R}$, let $y(x)$ be a solution of the differential equation

$\left(x^2-5\right) \frac{d y}{d x}-2 x y=-2 x\left(x^2-5\right)^2$ such that $y(2)=7$.

Then the maximum value of the function $y(x)$ is :
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4
JEE Advanced 2022 Paper 2 Online
Numerical
+3
-1
If $y(x)$ is the solution of the differential equation

$$ x d y-\left(y^{2}-4 y\right) d x=0 \text { for } x > 0, y(1)=2, $$

and the slope of the curve $y=y(x)$ is never zero, then the value of $10 y(\sqrt{2})$ is
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