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Mathematics · 2021

JEE Main · 17 March 2021, Shift 2 · Q68

Let y=y(x) be the solution of the differential equation cos x(3 sin x+ cos x+3) d y=(1+y sin x(3 sin x+ cos x+3)) d x, 0 ≤ x ≤ (π)/2, y(0)=0. Then,…

Let $\displaystyle y=y(x)$ be the solution of the differential equation $\displaystyle \cos x(3 \sin x+\cos x+3) \mathrm{d} y=(1+y \sin x(3 \sin x+\cos x+3)) \mathrm{d} x, 0 \leq x \leq \frac{\pi}{2}, y(0)=0$. Then, $\displaystyle y\left(\frac{\pi}{3}\right)$ is equal to :
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