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Differential Equations question

2022 · 27 Jul · Shift 1 · Q42
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  5. /2022 · 27 Jul · Shift 1 · Q42

Differential Equations question

2022 · 27 Jul · Shift 1 · Q42

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution curve of the differential equation sin⁡(2x2)log⁡e(tan⁡x2)dy+(4xy−42xsin⁡(x2−π4))dx=0\sin \left( {2{x^2}} \right){\log _e}\left( {\tan {x^2}} \right)dy + \left( {4xy - 4\sqrt 2 x\sin \left( {{x^2} - {\pi \over 4}} \right)} \right)dx = 0sin(2x2)loge​(tanx2)dy+(4xy−42​xsin(x2−4π​))dx=0, 0<x<π20 \lt x \lt \sqrt {{\pi \over 2}}0<x<2π​​, which passes through the point (π6,1)\left(\sqrt{\frac{\pi}{6}}, 1\right)(6π​​,1). Then ∣y(π3)∣\left|y\left(\sqrt{\frac{\pi}{3}}\right)\right|​y(3π​​)​ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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