# integrating factor

• June 9th 2008, 04:55 PM
i_zz_y_ill
integrating factor
Why can't I use an integrating factor when asked to find the complementaryt function of this?
dy/dt + 2y = e^-2t

I figured I either use an auxiliary eqn or I.F but the I.F doesn't work. Am i missing some conditions conditions to use an I.f or something thnx.

The I.f rule states in the form dy/dx +Py = Q
where P and Q are functions of x you can use the I.f
Since in this example 2 was essentially a function of x(i don't know how i just thought it is a constant) then arguably Q can just be a constant. Does that mean that i can use the I.f method for instance with dy/dx + 2y = 3??thnx
• June 9th 2008, 05:27 PM
topsquark
Quote:

Originally Posted by i_zz_y_ill
hy can't I use an integrating factor when asked to find the complementaryt function of this?
dy/dt + 2y = e^-2t

I figured I either use an auxiliary eqn or I.F but the I.F doesn't work. Am i missing some conditions conditions to use an I.f or something thnx.

Why won't it work? The integrating factor is $e^{2t}$, so your equation turns into
$\left ( e^{2t}y \right ) ^{\prime} = 1$

-Dan
• June 9th 2008, 05:41 PM
i_zz_y_ill
Yeah thanks, I was wondering though since in this stuation 2 is essentially a function of x if I had dy/dx + 3y = 4 couldn't this 4 essentially be a function of x in which case the method of I.f could be used. Giving y = 4/3 + c

What confuses me is why using an auxiliary eqn and a particular integral gives a more complicated genera solution i.e y = Qx + P + Ae^x

Furthermore what confuses me is that if 4 is just perceived as a function of x, what's to stop it being a constant if that makes sense function of y so that it can be taken to the LHS and the equation becomes homogenous i need of only a complementary function. In which case a general solution is y = Ae^-3x.

I've obviously presumed alot which is wrong!!!!I don't understand lol
• June 9th 2008, 05:48 PM
topsquark
Quote:

Originally Posted by i_zz_y_ill
Yeah thanks, I was wondering though since in this stuation 2 is essentially a function of x if I had dy/dx + 3y = 4 couldn't this 4 essentially be a function of x in which case the method of I.f could be used. Giving y = 4/3 + c

What confuses me is why using an auxiliary eqn and a particular integral gives a more complicated genera solution i.e y = Qx + P + Ae^x

Furthermore what confuses me is that if 4 is just perceived as a function of x, what's to stop it being a constant if that makes sense function of y so that it can be taken to the LHS and the equation becomes homogenous i need of only a complementary function. In which case a general solution is y = Ae^-3x.

I've obviously presumed alot which is wrong!!!!I don't understand lol

To solve $y' + 2y = e^{-2t}$ the general solution will be of the form $y(t) = Ae^{-2t} + Bte^{-2t}$, it does not have a linear term in it, if that's one of the things troubling you. (The A term is the solution to the homogeneous equation and the B term is the particular solution.)

The "4" can be taken to be a function of anything you desire. In the IF method we take it to be a function of x for convenience.

-Dan