Let y=Y(x) be the solution of the differential equation \frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x, x \\in (\frac{-\\pi}{2}, \frac{\\pi}{2}), such that Y(0)=1, then ______

Correct Answer is : y′(π4)−Y′(−π4)=π−√2

y(\frac{\\pi}{4})+Y(\frac{-\\pi}{4})={\\pi^{2}}/{2}+2

y'(\frac{\\pi}{4})+Y'(\frac{-\\pi}{4})=-\sqrt2

y(\frac{\\pi}{4})-Y(\frac{-\\pi}{4})=\sqrt2

y'(\frac{\\pi}{4})-Y'(\frac{-\\pi}{4})=\\pi-\sqrt2

AP EAPCET 25-Aug-2021 Shift 2 Paper

Section: Mathematics

Question : 80 of 160

Marks: +1, -0

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

+y‌tan‌x=2x+x2‌tan‌x,x∈ (

−π

), such that Y(0)=1, then ______

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