# Find the appropriate particular solution.

1. Nov 5, 2012

### Mukuro

1. The problem statement, all variables and given/known data

Find the appropiate particular solution form for

y''' - y'' - 12y' = x - 2x*e^-(3x)

DO NOT SOLVE FOR THE COEFFICIENTS

2. Relevant equations

Just guessing proper solution forms for nonhomogeneous second order D.Es with undetermined coefficients.

3. The attempt at a solution

Ok, as far as I know the solution in my mind was...

Ax + B + (Cx^2 + Dx) e ^(-3x)

The homogeneous solution was yc = c1 + c2e^(4x) + c3e^(-3x)
BUT

the solution is Ax + Bx^2 + (Cx + Dx^2)e^(-3x)

I don;t know why the B has x^2, and I don't know why D and C are switched around. I've been going at this for a good hour.

Last edited: Nov 5, 2012
2. Nov 5, 2012

### Zondrina

So you want a particular solution to $L[y] = x - 2xe^{-3x}$

So by solving the homogeneous system $L[y] = 0$, you get a solution yc(x).

Now, your particular solution, which I'll denote Y(x), cannot be the same as any of the solutions of the homogeneous system.

So since you've found your homogeneous solution to L[y] = 0 to be $y_c(x) = c_1 + c_2e^{-3x} + c_3e^{4x}$ your particular solution Y(x) cannot be equal to any of your terms y1, y2 or y3.

3. Nov 5, 2012

### Mukuro

I understand that e ^-3x is in the homogeneous solution, that's why it by x, however this still does not explain why C and D happen to be reversed or why b has an x^2 attached.

4. Nov 5, 2012

### Zondrina

You stated your idea for your particular solution was : Ax + B + (Cx^2 + Dx) e ^(-3x)

Notice you have e^(-3x) in your particular solution when I said that if a solution appeared in your homogeneous solution, then it could not be a particular solution.

5. Nov 5, 2012

### Mukuro

Oh I typed the actual solution wrong, which I'll have to fix.

The actual solution is ax + bx^2 (cx + dx^2) e^(-3x),

and I know that e^(-3x) is in the solution that is why I multiply it by and distribute it out.

x * (cx + b)*e^(-3x) is not equal to (cx +b)*e^(-3x).

6. Nov 5, 2012

### Zondrina

No no, since e^(-3x) appeared in your solution to the homogeneous system, it cannot appear in your formulation of a particular solution. So e^(-3x) CAN'T be part of the particular solution.

There's also a very simple fix to this and it has to do with the order of your equation.

7. Nov 5, 2012

### Mukuro

But the problem with that is, the book CLEARLY SAYS that e^(-3x) is apart of the particular solution, I have double checked two books, my own and my fellow classmates.

It also states that when this happens we can multiply the function by x to achieve Axe^(x) (standard example). This is what I have done.

8. Nov 5, 2012

### Zondrina

Okay so you know what you're doing, you're very close actually.

Why is multiplying by just x insufficient? Look at the order of your equation.

9. Nov 5, 2012

### Mukuro

Could you give me a hint as to what equation your referring to, my solution or the actual equation stated in the problem? I'm so very determined atm <.>

OK I recognize that it is a 3rd order equation... I do not however know what that does LOL. I usually deal with just 2nd orders.

10. Nov 5, 2012

### Zondrina

Since you have a third order equation, it may be the case that you must multiply by at most x3 to find your particular solution.

Also note, you can break your equation $L[y] = x - 2xe^{-3x}$ into two different problems :

$L[y] = x$ and $L[y] = -2xe^{-3x}$ and solve both of those instead of solving it all at once ( If it helps ).

11. Nov 5, 2012

### Mukuro

Well we don't have to solve it which I am happy about. However, I am still confused, even if I may have to multiply by at most x^3, it doesn't seem to explain the bx^2.

For instance the Ax + B is my guess to Ly = x, which I figure should be enough but seeing as it is a 3rd order equation how would I go about getting Ax + bx^2, I honestly have no idea how anyone could possibly guess that.

12. Nov 5, 2012

### Zondrina

You want a particular solution to : $L[y] = x - 2xe^{-3x}$

Split it into two cases : So if Y is the particular solution, it must be the sum of the solutions Y1 and Y2 of the equations :

Case one : $L[Y_1] = x$

Case two : $L[Y_2] = -2xe^{-3x}$

Find solutions to those ones because it's easier, then add them together. The case with x is not too bad since after you take a few derivatives... it's easy. Then just add the two solutions of those equations together to get Y.

EDIT : I'm almost positive you should have a theorem like this in your book?