Question

Need someone good at undetermined coefficients (Diff. Eqs.)

Answer 1

Suppose I have a differential equation in a form,
\(\large \color{blue}{(D-\alpha)^n[y]=e^{\alpha x}}\)

I know how to find the general solution,
\(\large \color{blue}{y_h=c_1e^{\alpha x}+c_2xe^{\alpha x}+\dots +c_nx^{n-1}e^{\alpha x} }\)

Or, I can write it as,
\(\large \color{blue}{\displaystyle y_h=e^{\alpha x}\cdot\sum_{i=1}^{n}c_i\cdot x^{i-1} }\)

but, what should my initial guess for \(\large \color{blue}{\displaystyle y_p }\) look like?

Answer 2

For example, if I have
\(\large \color{blue}{\displaystyle y'''-3y''+3y-1=e^x}\)

the auxiliary equation for the corresponding homogeneous differential equation quickly factors into,
\(\large \color{blue}{\displaystyle (r-1)^3=0}\)

So, my homogeneous solution is:
\(\large \color{blue}{\displaystyle y_h=c_1e^x+c_2xe^x+c_3x^2e^x}\)

And my question now is, how do I guess the (form of the) particular solution?

Answer 3

@freckles

Answer 4

so the particular solution would be
\[y=Ae^{x}\]
but this is already in the homog part
so we would tact on another x but that is all in there
and so is x^2
so I think the particular solution might be
\[y=Ax^3 e^x\]
but let me check... real quick
one sec

Answer 5

I think I should agree

Answer 6

The same idea with the second order (with r=1, and e^(1•x))
http://www.wolframalpha.com/input/?i=y%27%27-2y%27%2By%3De%5Ex

Answer 7

And the part. solution there involves quadratic

Answer 8

So, here it would be cubic by the same convention.
Thanks !!

Answer 9

I will solve that one (in my link wolfram) by hand, and if it matches, I will proceed to this 3rd order diff eq.

Answer 10

hey

Answer 11

I see one problem... I think

Answer 12

yes?

Answer 13

\[y'''-3y''+3y-1=e^x \text{ doesn't have characteristic equation } (r-1)^3=0 \\ \text{ you should actually be looking at it as } \\ y'''-3y''+3y=e^x+1 \\ \text{ the characteric equation should be } r^2-3r+3=0\]

Answer 14

I can't believe I wrote that!
Sorry, it should be
3y''' - 3y'' + 3y' - y =e^x

Answer 15

The point is:

what to do in case (r-a)^n=0 is an n-times repeating r=a, and
the f(t)=e^(at).

Answer 16

f(t), is my (and my textbook's) way of denoting the function (usually on Left hand side) that the differential equation = to.

Answer 17

but yeah for that corrected one the particular solution guess would be y=Ax^3e^x

Answer 18

so I think if you had (r-a)^n=0 as the characteristic equation
and f(t)=e^(at) then the particular solution guess would be
\[y=A t^n e^{at}\]

Answer 19

I hate the site!
It reloads me without me doing anything what so ever!

Yes, so in general in such case (1st thing in the post that I proposed),
the general guess would be

\(y_p=Ax^ne^{\alpha x}\)

Answer 20

yes it to me a few times a second ago

Answer 21

yes, you said it first, but you used a, and I used alpha :P
jk

Answer 22

Thanks for your clarification!
I will play with this stuff a little more, but nice that there are users like you who actually know!

Answer 23

Oh, I got a new hypothesis tested!

Answer 24

\(\large \color{blue}{(D-\alpha )^n[y]=e^{\alpha x}}\)

has a solution,
\(\large \color{blue}{y(x)=y_h+y_p}\)

where
\(\large \color{blue}{y_h=c_1e^{\alpha x}+c_2xe^{\alpha x}+\dots +c_nx^{n-1}e^{\alpha x} }\) as I said

\(\large \color{blue}{y_p=\frac{x^ne^{\alpha x}}{n!} }\)

Answer 25

smaller example
\[y''-6y'+9y=e^{3t} \\ \text{ characteristic equation } (r-3)^2=0 \implies r=3,3 \\ y_h=c_1 e^{3t}+c_2 te^{3t} \\ \text{ so } y_p=A t^2e^{3t}\]

\[y_p'=A2t e^{3t}+At^2 3e^{3t}=2A te^{3t}+3At^2e^{3t} \\ y_p''=2Ae^{3t}+2At3 e^{3t}+3A2te^{3t}+3At^23e^{3t}=9 At^2 e^{3t}+12Ate^{3t}+2Ae^{3t} \\ \\ \\ \\ \]



\[ y''_p-6y_p'+9y_p= \\ 9At^2e^{3t}+12Ate^{3t}+2Ae^{3t} \\ -12Ate^{3t}-18At^2e^{3t} \\ +9At^2e^{3t} \\ =2Ae^{3t} \\ 2Ae^{3t}=e^{3t} \implies 2A=1 \implies A=\frac{1}{2} \\ y_p=\frac{1}{2} t^2 e^{3t}\]

testing just for fun
i think this hypothesis does work

Answer 26

that wasn't a proof
just testing it to a small problem

Answer 27

Yes, an example is never a proof, but is a perfect disproof:)

Answer 28

I don't think I can actually prove this quite yet, although perhaps with anhilliator method that I haven't read yet.

Answer 29

I just glanced there, but won't go too too ahead.

Answer 30

oh you even guess the constant coefficient value of the particular solution which seems to test out good too

Answer 31

Yes, but I will now attempt to prove, or at least somewhat make more sense than what I made already.

Answer 32

so are you in a math class where you are making your own conjectures and proving them or is this just a fun experiment for you ?

Answer 33

\(\large \color{blue}{(D-a)^n[y]=e^{ax} }\)

using the annihilator method, (D-a annihilates e^(ax))

\(\large \color{blue}{(D-a)(D-a)^n[y] =0 }\)
\(\large \color{blue}{(D-a)^{n+1}[y] =0 }\)

\(\large \color{blue}{y=c_1e^{ax}+c_2xe^{ax}+\dots + c_{n+1}x^{n+1}e^{ax} }\)

Answer 34

N-th order diff. eq. is to have only N arbitrary constants, so therefore (C_(n+1)) is an undetermined coefficient, which I will solve for (if I can).

Answer 35

No I give up.

Answer 36

there might be some linear algebra approach, but not now ... ;(