# Method of undetermined coefficients, when to raise the guess

1. Jun 3, 2012

### Uku

Hello!

I have some examples of non-homogeneous ODEs to be solved by the undetermined coefficients method. Two from "Pauls math notes" page:

$y''+8y'+16y=e^{-4t}+(t^2+5)e^{-4t}$
The compsol. for this is:
$Y_{c}=C_{1}e^{-4t}+C_{2}te^{-4t}$
The first guess for a particular solution would be:
$Y_{p}=(At^2+Bt+C)e^{-4t}$
When multiplying the brackets with the exponent we get a solution which is in the complementary solution, thus we raise the order, indeed we have to do this twice. Pauls notes (link here) stress that only the part which is the mixed solution needs to be raised. In this example I would then raise only $Bt$ by $t^2$ and $C$ by $t^2$. But the correct form is $t^2(At^2+Bt+C)e^{-4t}$. Wolfram alpha verifies.

While on the other hand, of his other example:
$y''+3y'-28y=7t+e^{-7t}-1$
Complementary solution being $Y_{c}=C_{1}e^{4t}+C_{2}e^{-7t}$
The first guess at a particular solution:
$Y_{p}=At+B+Ce^{-7t}$
Now here the solutions match as well, meaning a raise of the particular matching exponential function:
$Y_{p}=At+B+Cte^{-7t}$ which is the correct guess.

Now... why in the first example is the whole polynomial multiplied by the $t^2$?

And another example:
$y^{(4)}-2y^{(3)}+y''=e^t+1$
The compsol. here is:
$Y_{c}=C_{1}+C_{2}+C_{3}e^{t}+C_{4}te^{t}$
Now, my first guess for a particular solution would be $Ae^t+B$
This is also a solution in the homogeneous equation, so I need to raise it, twice. But according to two sources the correct way of doing this is $At^{2}e^{t}+Bt^2$
Why not raise the function that is actually lapping, meaning $At^{2}e^{t}+B$

So.. that is the question.
Thank you,
U.

2. Jun 8, 2012

### algebrat

There is a form that the driving function (inhomogeneous part) should be in to apply thise methods. In some of your examples, the driving function is not of this form, so you break the problem up to find the two particular solutions corresponding to the two driving functions. However, for the third example, you have a fourth derivative so you'll have to be even more careful.

3. Jul 2, 2012

### Mandlebra

Note the repeated roots of the characteristic equation

4. Jul 5, 2012

### haruspex

I can make Paul's procedure consistent if it works like this:
- if a part of the PI matches a part of the CS then multiply it up by t
- if that makes it match another part of the CS, multiply that up too
- apply the above rules iteratively until it settles down