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Mathematics · 2025

JEE Main · 2 April 2025, Shift 2 · Q25

Let y=y(x) be the solution of the differential equation (d y)/(d x)+2 y sec^2 x=2 sec^2 x+3 tan x · sec^2 x such that y(0)=5/4. Then 12(y((π)/4)-…

Let $\displaystyle y=y(x)$ be the solution of the differential equation $\displaystyle \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y \sec ^2 x=2 \sec ^2 x+3 \tan x \cdot \sec ^2 x$ such that $\displaystyle y(0)=\frac{5}{4}$. Then $\displaystyle 12\left(y\left(\frac{\pi}{4}\right)-\mathrm{e}^{-2}\right)$ is equal to $\displaystyle \_\_\_\_$.
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.