Quick question on repeated roots when solving differential equations

[rwooduk]

say we have gone through the steps and have...

$(\lambda - 2)^{2}(\lambda ^{2}-9) = 0$

which we can write as...

$(\lambda - 2)(\lambda - 2)(\lambda ^{2}-9) = 0$

we have value for lambda of 2, 2, 3, -3

because we have a repeated root.

now, say we have

$(\lambda^{2} - 1)(\lambda + 1) = 0$

my question is would you have 2 values of -1 like the example above i.e. 1, -1, -1? or would you have values of lambda of just 1, -1? and if the former would you treat the -1 as a repeated root?

it becomes important when writing the solution because for a repeated root there an x in the second term with the exponential.

thanks in advance for any direction on this.

 

[HallsofIvy]

This isn't really a question about differential equations, it is a question about basic algebra.

Since $\lambda^2- 1= (\lambda- 1)(\lambda+ 1)$, which I am sure you already knew,
$(\lambda^2- 1)(\lambda+ 1)= (\lambda- 1)(\lambda+ 1)(\lambda+ 1)= (\lambda- 1)(\lambda+ 1)^2$
so, yes, -1 is a double root.

 

[rwooduk]

ahh, never thought of it that way! that's great, many thanks!