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Third order ODE, constant coefficients but inhomogeneous

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  • #1
fluidistic
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Homework Statement


Hello guys. I'm totally stuck at finding the solution to ##y'''-12y'+16y=32x-8##.


Homework Equations


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


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 [tex]\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[/tex]. 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. :smile:
 

Answers and Replies

  • #2
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Homework Statement


Hello guys. I'm totally stuck at finding the solution to ##y'''-12y'+16y=32x-8##.


Homework Equations


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


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.
First things first. Look at the characteristic equation for the homogeneous equation, which is r3 - 12r + 16 = 0. This factors very nicely.
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.
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.
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 [tex]\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[/tex]. 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. :smile:
 
  • #3
fluidistic
Gold Member
3,671
110
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).
Ah ok thanks, I was totally ignorant about this.
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.
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
fluidistic
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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
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Ah ok thanks, I was totally ignorant about this.

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.
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).
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.
 
  • #6
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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##.
That's higher degree than you need. Try a particular solution (of the nonhomogeneous problem) of yp = a0 + a1x.
 
  • #7
fluidistic
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That's higher degree than you need. Try a particular solution (of the nonhomogeneous problem) of yp = a0 + a1x.
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
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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+...
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
fluidistic
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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.
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.
 

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