# Homework Help: Differential equation: undetermined coeff.

1. Mar 22, 2007

### mrmotobiker

1. The problem statement, all variables and given/known data
find the general solution:

y''+2y'+y=2e^-t

2. Relevant equations

CE: r^2+2r+1=0
(r+1)^2 which means r1=r2=-1
c1e^-t+c2te^-t

3. The attempt at a solution

Y(t)=Ae^-t
Y'(t)=-Ae^-t
Y''(t)=Ae^-t
back into the equation:
Ae^-t-2Ae^-t+Ae^-t=2Ae^-t
0=Ae^-t
now im stuck! what am i doing wrong?! this is the "easiest" problem! isn't the A left there and that's how i find the GS?

the answer is supposed to be c1e^-t+c2tAe^-t+t^2Ae^-t

2. Mar 22, 2007

### Mindscrape

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?)

3. Mar 22, 2007

### mrmotobiker

but isn't your guess supposed to be Ae^-t because that's what fits. sorry. but not sure if i'm getting your question. thanks for the input.

4. Mar 22, 2007

### Mindscrape

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.

5. Mar 22, 2007

### mrmotobiker

ok i tried Ae^-t+Bte^-t as a guess. but it all canceled out still.

6. Mar 23, 2007

### Mindscrape

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

You need to try another.

7. Mar 23, 2007

### mrmotobiker

can i have a hint? =) i just tried Ate^-t from another book that says if the assumed solution is a multiplicative function of the homogenous solution and that it should be multiplied by t^m where m is the smallest integer possible. (did i do that right?)

8. Mar 23, 2007

### Mindscrape

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.

9. Mar 23, 2007

### mrmotobiker

ok, so i tried this:

Ate^-t+Bt^2 e^-t+C (not sure if i need the C)

but after all the algebra, B=1, and i'm not sure how to find A. am i on the right track? based on your explanation....i might be...

ahhhh i tried this like 6 times...

Last edited: Mar 23, 2007
10. Mar 23, 2007

### mrmotobiker

wait. A=C=0? so the solution is homogenous solution+this so it is:

c1e^-t+c2te^-t+t^2e^-t since B=1 -> 1te^-t?

11. Mar 23, 2007

### Mindscrape

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
$$y_p(t) = Ae^-t + Bte^-t + Ct^2e^-t$$

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

$$2e^-t = 2Ce^-t$$

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 $$y_p(t) = 0e^-t + 0te^-t + 1t^2 e^-t$$.

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

12. Mar 23, 2007

### Mindscrape

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. :)

13. Mar 23, 2007

### mrmotobiker

ok great, thanks! you've been a lot of help. but i guess i still made a mistake because my guess was Ate^-t+Bt^2 e^-t+C instead of Ae^-t+Bte^-t+Ct^2e^t. so just to confirm, the guess should be At^-t + Bt^-t + Ct^2e^-t? i thought it was just to multiply the homogeneous solution by a factor of t...sorry...taking a while to digest, this stuff is still really new.

14. Mar 23, 2007

### Mindscrape

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.

15. Mar 23, 2007

### HallsofIvy

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.

16. Mar 23, 2007

### mrmotobiker

great, thanks everyone. ok i'm just practicing problems now since i have a midterm coming up. i'm using schaum's outlines: differential equations. here's the problem:

y''-2y+y=t^2-1

CE: (r-1)^2, r1=r2=1
yh=c1e^t+c2te^t

GS=yh+yp

my guess: yp=At^2+Bt+C
yp'=2At+B
yp''=2A

(question: since it's a repeated root, should i multiply my guess by a factor of t?)

so after all the algebra, i got yp=t^2-4t-11

the book solution is yp=t^2+4t+5; GS=c1*e^t+c2*te^t+t^2+4t+5

did i make an algebra mistake? i went through and looked, it doesn't look like it. did i choose my guess wrong? why?

17. Mar 23, 2007

### HallsofIvy

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

Since your solution to the entire equation does not involve et, 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!

Last edited by a moderator: Mar 26, 2007
18. Mar 24, 2007

### mrmotobiker

i just realized. for the first problem, my professor said resonance analysis must be performed. how do you know when you need to do this?

19. Mar 24, 2007

### Mindscrape

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

20. Mar 24, 2007

### mrmotobiker

umm well he said resonance analysis is make r = alpha + beta*i s=# of times r appears as a root of CE. the equation he keeps using is:

y=t^s * e^(alpha*t) * (An * cos(beta*t) + Bn * sin(beta*t))

i'm not sure when to apply this, but one of my classmates said i am supposed to perform this analysis to get Ate^-t as a guess

21. Mar 25, 2007

### mrmotobiker

is that an easier...or more precise way of making a "guess?"

22. Mar 25, 2007

### Mindscrape

Ah, I see. The equation for the "resonance analysis," is merely a simplified form of what you would find through normal second order ODE methods. If you have an equation of the form ay'' + by' + ky = f(t), and the coefficients for 'a,' 'b,' and 'k' produce complex solutions then the homogeneous will be the form that you listed (except for the t^s out front, for which I am not sure of where it came in).

I would say to memorize the solution (what you have without the t^s) for an underdamped oscillator for your test, since deriving it will waste time, but you should definitely know how to get to that point. Or if you are allowed to leave the solution in it's complex form then just use the normal method of finding the characteristic roots, aka eigenvalues.

Anyway, the second problem you listed is a case of a critically damped oscillator, and will not have the solution you are talking about. The critically damped solution is simply an example of a repeated real root.

23. Mar 26, 2007

### mrmotobiker

ok that makes more sense. thanks again.

24. Mar 26, 2007

### mrmotobiker

oh, the professor said the t^s is there just in case if there is "resonance." the s is the number of times the root appears in the CE; but i am really unsure why did he put the t^s, i haven't seen this in the books i'm using.