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JEE MainMathematicsModerateDifferential equation

The solution of differential
equation
dydx=y2y(siny2+cosy2)−x \frac{dy}{dx} = \frac{y}{2y\left( siny^{2} + cosy^{2} \right) - x}\ is.
vody lehdj.k dk gy

dydx=y2y(siny2+cosy2)−x\frac{dy}{dx} = \frac{y}{2y\left( siny^{2} + cosy^{2} \right) - x}

Show answer and solution
dxdy+xy=2(siny2+cosy2)\frac{dx}{dy} + \frac{x}{y} = 2\left( siny^{2} + cosy^{2} \right)
 I.F.
= e∫1ydy=y= \ e^{\int\frac{1}{y}dy} = y
Solution if gy gksxk;fn

xy=∫2y(siny2+cos⁡2y2)dy=−cosy2+siny2+Cxy = \int 2y\left( siny^{2} + \cos^{2}y^{2} \right)dy = - cosy^{2} + siny^{2} + C

Source: Grass-G / M-21 Mathematics Paper 19 Dec Hindi Eng.docx

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