Question

I have two questions and both are differential equations problems on the topic of undetermined coefficients:

1) y''(x)+y(x)=2^x

2) y''-2y'+y=7e^(t)cos(t)


Help would be most appreciated!

Answer 1

What are your initial conditions?

Answer 2

Oh I am sorry, the first one is asking for a particular solution to the differential equation and the second is asking for the same thing but without solving for the coefficients.

Answer 3

Okay--we just covered this in my DiffEQ class. Find the homogenous solution first...

y"(x)+y(x)=0

m^2+m=0->m=0, m=-1

Which means it's e^0+e^(-x)...that's how you start the first one. I'm realizing at this point that I really haven't practiced these enough to help. I might be able to do something more specific--or I'll look at my book in a few minutes.

Answer 4

Umm you had me until e^0+e^(-x)..
Haha okay that's fine.

Answer 5

I know how to do the second one with linear algebra instead of undetermined coefficients. But I doubt you were taught that.

Answer 6

Nope :/ but thank you anyways.

Answer 7

Well what's your professor expect if not by undetermined coefficients? Or do they just mean pick what you would choose as your guess and just not solve it? In that case that's easy and I can help show you why.

Answer 8

For which one?

Answer 9

The second one. I can help you solve both if you need help though.

Answer 10

That would be wonderful if you can.

Answer 11

The first one just needs to be rewritten:
\[2^x=e^{\ln2*x}\]

So now you just say

\[y=Ae^{\ln2*x}\]

Answer 12

Okaay..

Answer 13

The next one, since he doesn't want you to find the coefficients I think that just means that it's just a really long problem, so just make the correct guess. How do you know though?

Well you have a scalar multiple of e^t*cos(t). You know the derivative by using the product rule will give you some other scalar multiple of e^t*cos(t) and e^t*sin(t). The derivative of the e^t*sin(t) will also just give you a scalar multiple of e^t*cos(t) and e^t*sin(t) so you now can see how your guess should just be this added together with a different coefficient on each.

\[y=Ae^t \cos t+Be^t \sin t \]

Answer 14

Ask questions you seem lost.

Answer 15

Okay so for the first one (because I actually have to solve for the coefficients for that one), I just take the second derivative of that? But why do I feel like I will just be going in circles? Because the e and the ln will cancel which will just give me y=A2^(x), right?

And for the second one, that is what I had but then doesn't it matter that there is a double root for the auxiliary equation? (which is just r^2-2r+1=0)

Answer 16

I do not have the answer for the second one but apparently the answer for the first one is: y(x)=[(ln2)^2+1]^(-1)2^x

Answer 17

I'm just showing that if it looks like that it might be nicer. It was just a cosmetic choice to make it absolutely obvious that the derivative of it will just be a scalar multiple of itself. If it seems confusing, just use y=A*2^x

Answer 18

Okay

Answer 19

That doesn't really answer the second question though. :p

Answer 20

So wait, are you saying you understand and are done with question 1? Let's do things one at a time. Where are you on 1? Done?

Answer 21

I am done with question 1 :)

Answer 22

If you can explain to me why it doesn't matter that the auxiliary equation has a double root, then I will be golden.

Answer 23

Because I got what you original had put for number 2, but without considering the left side.

Answer 24

Hmm well I'm thinking and since you're looking for a particular solution it shouldn't matter to us. If you are wanting the general solution then perhaps you should care... Even then I'm sort of doubtful. It's been too long I think.

Answer 25

haha okay well th

Answer 26

thank you*

Answer 27

You don't have to write anymore :) I am good. Thank you for your help!

Answer 28

You might try calculating the wronskian or something. lol good luck.