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Differential Equations question

2019 · 10 Apr · Shift 1 · Q26
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  5. /2019 · 10 Apr · Shift 1 · Q26

Differential Equations question

2019 · 10 Apr · Shift 1 · Q26

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If y = y(x) is the solution of the differential equation dydx=(tan⁡x−y)sec⁡2x{{dy} \over {dx}} = \left( {\tan x - y} \right){\sec ^2}xdxdy​=(tanx−y)sec2x, x∈(−π2,π2)x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)x∈(−2π​,2π​), such that y (0) = 0, then y(−π4)y\left( { - {\pi \over 4}} \right)y(−4π​) is equal to :
  1. A
    12−e{1 \over 2} - e21​−e
  2. B
    e−2e - 2e−2
  3. C
    2+1e2 + {1 \over e}2+e1​
  4. D
    1e−2{1 \over e} - 2e1​−2
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