Verified Solution • 2025–26 Session moderate

QJEEM19Ma07dc: JEE Main 2019 Mathematics - Class 12 JEEM19Ma07dc Solution: If     and  …

If   dydx+3cos⁡2xy=1cos⁡2x,  x∈(−π3,π3){{dy} \over {dx}} + {3 \over {{{\cos }^2}x}}y = {1 \over {{{\cos }^2}x}},\,\,x \in \left( {{{ - \pi } \over 3},{\pi \over 3}} \right)   and   y(π4)=43,y\left( {{\pi \over 4}} \right) = {4 \over 3},   then   y(−π4)y\left( { - {\pi \over 4}} \right)    equals -

Multiple Choice Options

  • A 13+e6{1 \over 3} + {e^6}
  • B 13{1 \over 3}
  • C 13{1 \over 3} + e3
  • D −- 43{4 \over 3}

Step-by-Step Verified Solution

dydx+3sec⁡2x.y=sec⁡2x{{dy} \over {dx}} + 3{\sec ^2}x.y = {\sec ^2}x

I.F. = e3∫sec⁡2xdx=e3tan⁡x{e^{3\int {{{\sec }^2}xdx} }} = {e^{3\tan x}}

or    y.e3tan⁡x=∫sec⁡2x.e3tan⁡xy.e{}^{3\tan x} = \int {{{\sec }^2}x.{e^{3\tan x}}}

or    y.e3tan⁡x=13e3tan⁡x+Cy.{e^{3\tan x}} = {1 \over 3}{e^{3\tan x}} + C

Given

y(π4)=43y\left( {{\pi \over 4}} \right) = {4 \over 3}

∴\therefore     43.e3=13e3+C{4 \over 3}.{e^3} = {1 \over 3}{e^3} + C

∴\therefore    C = e3

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