# Homework Help: General solution of third order DE

1. Aug 19, 2010

### beetle2

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

y'''-3y'+2y=0

initial conditions y(0)=0, y'(0)=1,y''(0)=1

2. Relevant equations

Assume $y=e^{rt}$

3. The attempt at a solution

By the substitution I'm left with

$r^3-3r+2=0$

which gives me the roots of -2 and 1.

my question is a lot of times with this type of question I can get three roots and get a genral solution of

$y=c_1e^{rx}+c_2e^{rx}+c_3e^{rx}$

this time my solution is

$y=c_1e^{-2x}+c_2e^{t}$
therefore
$y'=c_1-2e^{-2x}+c_2e^{t}$
and
$y''=c_14e^{-2x}+c_2e^{t}$

because I've only got the two constants I solved for $c_1=\frac{-1}{3}$ and $c_2=\frac{1}{3}$

The problem is when I substitue back into the solutions for y,y',y''

I'm getting

initial conditions y(0)=0, y'(0)=1 which is good but I'm getting y''(0)=-1.

can some please let me know where I'm going wrong?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 19, 2010

### rock.freak667

r3-3r+2=0 should give you three solutions.

3. Aug 19, 2010

### vela

Staff Emeritus
If you have a third-order differential equation, you need three linearly independent solutions. You only have two because one of the roots is repeated. What do you usually do when you have a repeated root? You do the same thing here.

4. Aug 19, 2010

### beetle2

So I have the roots of -2 ,1 and the repeated root 1.
forgive my ignorance but I don't know what to do with a repeated root

5. Aug 19, 2010

### vela

Staff Emeritus
Did you check your textbook? I'm sure it discusses what to do in that case. Here's a quick summary to jog your memory.

http://www.sosmath.com/diffeq/second/constantcof/constantcof.html

That page talks about second-order equations, but it applies to higher-order equations as well.

6. Aug 19, 2010

### beetle2

Hi Vela,

I see that pesky extra x sneaking in there when r1=r2.

So my general solution should be

$y=c_1e^{-2x}+c_2e^{x}+c_2xe^{x}$

Last edited: Aug 19, 2010
7. Aug 19, 2010

### vela

Staff Emeritus
Yes, and now with three linearly independent solutions, you can get a consistent solution for the constants that will satisfy the initial conditions.

8. Aug 22, 2010

### beetle2

Hi Guys,
using the general solution I found the following.

$y=c_1e^{-2x}+c_2e^{x}+c_3xe^{x}$
$y'=c_1-2e^{-2x}+c_2e^{x}+c_3e^{x}+xe^{x}$
$y''=c_14e^{-2x}+c_2e^{x}c_32e^{x}+xe^{x}$

Using the coefficients and initial conditions in a matrix.

$$\left( \begin{array}{cccc} 1 & 1 & 1 & 0 \\ -2 & 1 & 1 & 1\\ 4 & 1 & 2 & 1\end{array} \right)$$

I solved for the row reduced echelon matrix.

$$\left( \begin{array}{cccc} 1 & 0 & 0 & -\frac{1}{3} \\ 0 & 1 & 0 & -\frac{5}{3}\\ 0 & 0 & 1 & 2\end{array} \right)$$

which gives the solution of

$y=-\frac{1}{3}e^{-2x}+-\frac{5}{3}e^{x}+2xe^{x}$

which satisfies

$y(0)=0$
$y'(0)=1$
$y''(0)=1$

thanks for all your help guys.

It was a pretty challenging problem!