math problem

Verify that y=sinxcosx-cosx is a solution of the initial-value problem
y'+(tanx)y=(cos^2)x y(0)=-1 on the interval -pi/2<x<pi/2
y'=dy/dx=sinx(-sinx)+cosx(cosx)+sinx=cos^2x-sinx^2x+sinx
So
LHS
=y'+(tanx)y
=cos^2x-sinx^2x+sinx+tanx(sinxcosx-cosx)
=cos^2x-sinx^2x+sinx+sin^2x-sinx
=cos^2x
=RHS
Also
y(0)
=sin0*cos0-cos0
=-1
satisfy the condition
We verify that y=sinxcosx-cosx is a solution fo the initial-value problem