Y'-(1/tgx)y=(sinx)^2....y(pi/2)=2....can someone solve this? :)

1 Answer
May 10, 2018

#y = sin x ( C - cos x)#

Explanation:

#y'- cot x \ y=sin^2x#

Integrating factor:

  • #exp( int - cot x \ dx ) = exp(- ln sin x + C) = C csc x #

#y' csc x- csc x cot x \ y=sin^2x csc x#

#(y csc x)^' =sin x#

#y csc x =- cos x + C#

#y =- cos x sin x + C sin x = sin x ( C - cos x)#