Solving a Separable ODE: y'+ytanx=cosx with Initial Condition y(0)=1

[eeriana]

Homework Statement

y'+ytanx = cos x y(0)=1

Homework Equations

The Attempt at a Solution

We are studying separable ode's and integrating factor right now, I am a little confused... If someone could steer me in the right direction, it would be greatly appreciated... This is what I have so far:

P= tanx
\intP = -ln|cosx|
\mu=e^{}-lncosx
\mu= 1/cosx

(1/cosx*y)' =\int1/cosx cosx

and this is where I get stuck... am I even on the right track?

Thanks

Eeriana

 

[Hootenanny]

You are on the right track and have calculated the IF correctly and written the complete differential correctly. Note that you can simplify the integrand...

 

[eeriana]

But isn't 1/cosx * cosx = 1 Or am I having an algebraic malfunction?

 

[Hootenanny]

But isn't 1/cosx * cosx = 1 Or am I having an algebraic malfunction?

Nope you are indeed correct.

 

[eeriana]

I thought I was doing something wrong... hmmm..now I am going to see if I can finish it!

Thanks for the help