Homework Help: Differential equation: undetermined coeff.

Have you figured out the homogeneous solution yet? Do that and see if your problem pops out. (Hint: What do you think will happen if you guess the same form as your homogeneous solution?)

 

Yes, typically that would be a good guess at the particular solution, but essentially what happens is that the solution space (general solution) is the superposition of the homogeneous solution and the particular solution. If you add a constant multiple of the homogeneous solution then that will get you nowhere. In other words, if your guess at the particular solution has the same form as the homogeneous solution then you will not find the other linear function in the superposition of the general solution because L1[x] + cL1[x] doesn't really mean anything.

 

Of course it did. The homogeneous solution is c1e^-t + c2te^-t. Your guess is exactly the same.

You need to try another.

 

Alright, you were on the right track by continually adding coefficients and higher orders of t.

Did you know that you could solve an equation that only needed Ae^-t with as many higher order terms as you wanted?
Zt^n e^-t + Yt^n-1 e^-t + ... + Bt e^-t + Ae^-t would give the same thing for an equation of say y'' + y = 10e^-t as guessing the lowest order term Ae^-t.

 

Yep, you got it. It seems like you are not really sure why A and C are zero though.

So if you guess the particular solution
[tex]y_p(t) = Ae^-t + Bte^-t + Ct^2e^-t[/tex]

You will find a huge mess in your particular solutions. After a lot of algebra, which you have to be careful with and really just take it slow, you will find that

[tex] 2e^-t = 2Ce^-t[/tex]

multiply the equation by e^t and C = 1.

The other coefficients are completely gone, and so they must be zero.

Thus, the particular solution will be [tex]y_p(t) = 0e^-t + 0te^-t + 1t^2 e^-t [/tex].

The general solution is the superposition of the homogeneous and particular.

 

Ah crap, I forgot brackets. Just note that every time you see a e^-*t I really meant to do e^(-t). Don't really want to fix it. :)

 

Yeah, you should have had my guess, which is really (A + Bt + Ct^2)e^-t if that helps you see it a little clearer. You just have to make sure your guess at the particular solution doesn't copy the homogeneous solution because it will not have the range (solution space) as the homogeneous.

Say you had a homogeneous solution of c1e^t + c2t^2e^t + c3t^3e^t + c4t^4e^t then your particular would have to have one higher order than the homogeneous y_p = (A + Bt +Ct^2 + Dt^3 + Et^4 + Ft^5)e^-t.

 

You don't need all of (A+ Bt+ Ct²)e^-t. Since both e^-t and te^-t are solutions to your homogeneous equation, try y= At²e^-t.

 

I assume you mean y"- 2y'+ y= t^2- 1

Since your solution to the entire equation does not involve e^t, no.

yp"= 2A
yp'= 2At+ B so -2yp'= -4At- 2B
yp= At²+ Bt+ C
yp"- 2yp'+ yp= 2A- 4At- 2B+ At²+ Bt+ C
= At²+ (B-4A)t+ (2A- 2B+ C)= t²- 1
so we must have A= 1, B- 4A= 0, 2A- 2B+ C= -1.
B- 4(1)= 0 so B= 4, 2(1)-2(4)+ C= -1 so C= -1+ 8- 2= 5

I suspect that you lost track of signs!

 

What do you, and your professor, mean by resonance analysis?