Question

y.sin2x.dx-(1+y^2+cos^2x)dy=0​

Answer 1

Step-by-step explanation:

We have,

$y \sin(2x) dx - (1 + {y}^{2} + \cos^{2} (x) )dy = 0$

$\implies \: y \sin(2x) dx - \cos ^{2} (x) dy = (1 + {y}^{2} )dy$

$\implies - ( \cos^{2} (x) dy - y \sin(2x) dx) = (1 + {y}^{2} )dy$

$\implies \: - d( y.\cos^{2} (x) ) = (1 + {y}^{2} )dy$

Integrating both sides,

$\implies \: \int - d( y.\cos^{2} (x) ) = \int (1 + {y}^{2} )dy$

$\implies \: y \cos^{2} (x) = y + \frac{ {y}^{3} }{3} + c$