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 \leq x \leq \frac{\pi}{2}, y(0)=0. Then, y(\frac{\pi}{3}) is equal to [17 Mar 2021 Shift 2]

Correct Answer is : 2loge(2√3+1011)

2 \log _{e}(\frac{2 \sqrt{3}+9}{6})

2 \log _{e}(\frac{2 \sqrt{3}+10}{11})

2 \log _{e}(\frac\sqrt{3}+7}{2})

2 \log _{e}(\frac{3 \sqrt{3}-8}{4})

JEE Main 17 March 2021 Shift 2 Solved Paper

Section: Mathematics

Question : 63 of 90

Marks: +1, -0

Let y=y(x) be the solution of the differential equation cos‌x(3‌sin‌x+cos‌x+3)‌d‌y= [1+y‌sin‌x(3‌sin‌x+cos‌x+3)]dx 0≤x≤‌

,y(0)=0. Then, y(‌

) is equal to

[17 Mar 2021 Shift 2]

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