Struggling with Differential Equations

1. Feb 15, 2010

jasonbot

1. The problem statement, all variables and given/known data

Hi, I'm struggling with a differential equations:

y'=(1-y)cosx

2. Relevant equations

3. The attempt at a solution

y'=cosx-ycosx

p(x)=e^(∫-cosxdx)
=e^-sinx

y'e^-sinx=e^(-sinx)cosx-e^(-sinx)ycosx

then I get confused because of the derivative.

I thought it could be
y'e^(-sinx)=d/dx(e^(-sinx)y) but that clearly doesnt work...

2. Feb 15, 2010

CompuChip

Separation of variables might be a good approach here... you can show that the solution is given by

$$\int \frac{1}{1 - y} \, dy = \int \cos(x) \, dx$$

and both of the integrals are well calculable.

3. Feb 15, 2010

vela

Staff Emeritus
If you want to use an integrating factor, you need to move the y's to the same side.

$$y'+(\cos x)y = \cos x$$

When you multiply through by the integrating factor p(x), the LHS becomes the derivative of p(x)y(x).

$$\mbox{LHS} = p(x)y'(x)+p'(x)y(x) = [p(x)y(x)]'$$

That's where the y' term goes.

4. Feb 15, 2010

jasonbot

Gah! My bad :( I feel quite stupid now... The exercise is on integrating factors and I didn't see that
Thanks vela for clearing up integrating factors for me :D

so to continue...

∫1/(1-y)dy=∫cosxdx

-ln(1-y)=sinx+c

1-y=Ae^(-sinx), A=e^c

-y=Ae^(-sinx)-1
y=-Ae^(-sinx)+1

which according to the solution is correct. Thanks guys! It's actually an IVP but I'll just leave that out since it's easy.

5. Feb 15, 2010

CompuChip

Note that what you are doing now is separation of variables.
To use integrating factors, you should proceed along vela's lines.

6. Feb 15, 2010

jasonbot

I did note that, it's just easier this way. Thanks CompuChip