differential equations

jeca86

Junior Member
Joined
Sep 9, 2005
Messages
62
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?
 
It's a first order equation, so you only need the first derivative. Substitute your expression for y' into the original d.e. and use trig identities.
 
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