JEE Main 2025 (Online) 3rd April Evening Shift | Differential Equations Question 5 | Mathematics

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

$\frac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x, y(0)=\frac{1}{3}+e^3$. Then $y\left(\frac{\pi}{4}\right)$ is equal to :

$\frac{4}{3}$

$\frac{2}{3}+e^3$

$\frac{4}{3}+e^3$

$\frac{2}{3}$

JEE Main 2025 (Online) 3rd April Morning Shift

MCQ (Single Correct Answer)

-1

Let $g$ be a differentiable function such that $\int_0^x g(t) d t=x-\int_0^x \operatorname{tg}(t) d t, x \geq 0$ and let $y=y(x)$ satisfy the differential equation $\frac{d y}{d x}-y \tan x=2(x+1) \sec x g(x), x \in\left[0, \frac{\pi}{2}\right)$. If $y(0)=0$, then $y\left(\frac{\pi}{3}\right)$ is equal to

$\frac{4 \pi}{3}$

$\frac{2 \pi}{3}$

$\frac{2 \pi}{3 \sqrt{3}}$

$\frac{4 \pi}{3 \sqrt{3}}$

JEE Main 2025 (Online) 29th January Evening Shift

MCQ (Single Correct Answer)

-1

If for the solution curve $y=f(x)$ of the differential equation $\frac{d y}{d x}+(\tan x) y=\frac{2+\sec x}{(1+2 \sec x)^2}$, $x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right), f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{10}$, then $f\left(\frac{\pi}{4}\right)$ is equal to:

$\frac{5-\sqrt{3}}{2 \sqrt{2}}$

$\frac{4 - \sqrt{2}}{14}$

$\frac{9\sqrt{3} + 3}{10(4 + \sqrt{3})}$

$\frac{\sqrt{3} + 1}{10(4 + \sqrt{3})}$

JEE Main 2025 (Online) 29th January Morning Shift

MCQ (Single Correct Answer)

-1

Let y = y(x) be the solution of the differential equation :

$\cos x\left(\log _e(\cos x)\right)^2 d y+\left(\sin x-3 y \sin x \log _e(\cos x)\right) d x=0$, x ∈ (0, $\frac{\pi}{2}$ ). If $ y(\frac{\pi}{4}) $ = $-\frac{1}{\log_{e}2}$, then $ y(\frac{\pi}{6}) $ is equal to :

$\frac{2}{\log_{e}(3)−\log_{e}(4)}$

$-\frac{1}{\log_{e}(4)}$

$\frac{1}{\log_{e}(4)−\log_{e}(3)}$

$\frac{1}{\log_{e}(3)−\log_{e}(4)}$

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