# Third order ODE, constant coefficients but inhomogeneous

1. Nov 30, 2012

### fluidistic

1. The problem statement, all variables and given/known data
Hello guys. I'm totally stuck at finding the solution to $y'''-12y'+16y=32x-8$.

2. Relevant equations
Variation of parameters once I'm done with the general solution to the homogeneous ODE.

3. The attempt at a solution
First I want to solve the homogeneous ODE $y'''-12y'+16y=0$ and then use variation of parameters to solve $y'''-12y'+16y=32x-8$ or maybe I'll propose a solution such as a polynomial of degree 4, $P_4(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ to find the particular solution of the ODE.
So, dealing with the homogeneous ODE, I'm totally stuck. I tried to get the characteristic equation, it is worth $r^3-12r+16=0$ so it's not a piece of cake to get the 3 values of r which satisfy that equation. I could not do it.
I tried to seek a solution in terms of infinite series and since there's no singular point I sought for a solution of the type $\phi (x)=\sum _{n=0}^\infty a_n x^n$. Which eventually lead me to $$\sum _{n=0}^\infty n(n-1)(n-2)a_n x^{n-3}-12\sum _{n=0}^\infty a_n n x^{n-1}+16 \sum _{n=0}^\infty a_n x^n=0$$. I did not proceed further in that direction because from here I notice that I'll get $a_{n+3}$ in terms of $a_{n+1}$ and $a_n$ which turns out to be incredibly messy. I was expecting to get a single coefficient in terms of another single coefficient.
This is where I ran out of ideas. I'm looking for any idea guys to tackle this ODE. Thank you.

2. Nov 30, 2012

### Staff: Mentor

First things first. Look at the characteristic equation for the homogeneous equation, which is r3 - 12r + 16 = 0. This factors very nicely.
It actually was a piece of cake to factor. If there are rational factors, they have to be in the set {±1, ±2, ±4, ±8, ±16} (see Rational Root Theorem).

I used synthetic division to look for a factor and found one right away. If you don't know synthetic division, you can use polynomial long division. Synthetic division is pretty much the same as poly. long division, but eliminates a lot of the writing.

3. Nov 30, 2012

### fluidistic

Ok so I've found out that r=2 satisfies the char. equation. So I wrote the char. equation as $(r-2)(r-a)(r-b)=r^3-12r+16$, I expanded the left hand-side and equated the coefficients. I reached that the 3 roots are $r_1=-4$, $r_2=2$ and $r_3=2$.
So I guess that the solution to the homogeneous equation is of the form $\phi (x)=Ae^{-4x}+Be^{2x}+Cxe^{2x}$. I'll try now to solve the non homogeneous ODE.
Thanks so far.

4. Nov 30, 2012

### fluidistic

I proposed a solution of the form $\phi _p (x)=a_0+a_1x+a_2x^2+a_3 x^3 +a_4 x^4$. Plugging that back (and its first and third derivative) into the non homogeneous ODE I reached that $a_0=1$ and $a_1=2$ so that the particular solution is $\phi _p (x)=1+2x$. So that the general solution to the ODE is $y(x)=Ae^{-4x}+Be^{2x}+Cxe^{2x}+1+2x$.
I'll try to use variation of parameters to get the same result as it's a weakness of mines.

5. Nov 30, 2012

### Staff: Mentor

That looks like the hard way. After I found that r = 2 was a root of the char. equation, the other factor was r2 + 2r - 8, which factors easily enough to (r + 4)(r - 2).

6. Nov 30, 2012

### Staff: Mentor

That's higher degree than you need. Try a particular solution (of the nonhomogeneous problem) of yp = a0 + a1x.

7. Nov 30, 2012

### fluidistic

Yeah I noticed this once I reached the solution. But how can you know this beforehand? My assumption made sense to me because I have to take the 3rd derivative of a polynomial and there's a chance that a term with "x" survives afterward since the ODE is y'''+...=32x+...

8. Nov 30, 2012

### Staff: Mentor

With all things being equal, if you have a polynomial of degree n on the right side, your particular solution should be a polynomial of degree n. Your DE has a y term in it, which will contribute the terms you need.

By "all things being equal", what I mean is that r = 0 is not a solution of the characteristic equation. I don't want to say much more than that, since the explanation can get pretty long.

9. Nov 30, 2012

### fluidistic

Ah ok thanks for the information. Good to know, this will save me time.
Anyway I tried variation of parameters and abandonned after calculating the second derivative of the assumed particular solution, it's very long and the 3rd derivative sounds like a nightmare in terms of number of terms. I'll stick with variation of parameters only for ODE of order 2 unless I'm really stuck. At least I reached the answer here, via another method.