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
so i have the first and second derivs.
y'=-(sin^2)x+(cos^2)x+sinx
y"=-4cosxsinx+cosx
Is this right? What do I do next?
y'+(tanx)y=(cos^2)x y(0)=-1 on the interval -pi/2<x<pi/2
so i have the first and second derivs.
y'=-(sin^2)x+(cos^2)x+sinx
y"=-4cosxsinx+cosx
Is this right? What do I do next?