# Help with Third-order linear ordinary differential equation

Printable View

• Dec 4th 2009, 11:45 AM
cYn
Help with Third-order linear ordinary differential equation
OK here is my problem:

$y''' - y'' - 2y' = 0$

And these are my steps

$m^3 - m^2 - 2m = 0$
$m(m^2 - m - 2) = 0$
$m(m + 1)(m - 2) = 0$
$m = -1, 0, 2$

*C1, C2, C3 = Csub1, Csub2, Csub3

y = C1e^0x + C2e^2x + C3e^-x
solution: y = C1 + C2e^2x + C3e^-x

but when checking the problem on Wolfram Alpha, the solution is:

y = C1 + (1/2)C2e^2x + C3(-e^-x)

Is there a step I'm missing?
• Dec 4th 2009, 01:30 PM
Jester
Quote:

Originally Posted by cYn
OK here is my problem:

$y''' - y'' - 2y' = 0$

And these are my steps

$m^3 - m^2 - 2m = 0$
$m(m^2 - m - 2) = 0$
$m(m + 1)(m - 2) = 0$
$m = -1, 0, 2$

*C1, C2, C3 = Csub1, Csub2, Csub3

y = C1e^0x + C2e^2x + C3e^-x
solution: y = C1 + C2e^2x + C3e^-x

but when checking the problem on Wolfram Alpha, the solution is:

y = C1 + (1/2)C2e^2x + C3(-e^-x)

Is there a step I'm missing?

These answers are the same. It might help if you choose different constant names like

$y = k_1 + k_2 e^{2x} + k_3 e^{-x}$

so you can choose your constants to recover Wolfram's.
• Dec 4th 2009, 01:34 PM
cYn
Quote:

Originally Posted by Danny
These answers are the same. It might help if you choose different constant names like

$y = k_1 + k_2 e^{2x} + k_3 e^{-x}$

so you can choose your constants to recover Wolfram's.

Ahh, thank you.