- #1
Mozart
- 106
- 0
When given
y'' + ay' + by + c = 0
The characteristic equation is:
λ^2 + aλ + b = 0
Now I'm making a substitution of e^(λx) to get
[(λ^2 + aλ + b)][e^(λx)] = 0
double root : [(λ - alpha)^2] [e^(λx)] = 0
My question now is that if I were to do the same thing with the following equation would it follow that
y''' + ay'' + by' + cy +d = 0
Characteristic equation λ^3 + aλ^2 + bλ + c
then making the same substitution ( λ^3 + aλ^2 + bλ + c ) e^(λx)
NOW HERE IS MY QUESTION: Can I write the above as
[(λ - alpha)^3][e^(λx)]
or do I have to write it a different way. Thank you very much.. I know this doesn't really have much to do with the solving of differential equations but it would be very helpful if I knew that that was correct for my assignment.
Thanks.
y'' + ay' + by + c = 0
The characteristic equation is:
λ^2 + aλ + b = 0
Now I'm making a substitution of e^(λx) to get
[(λ^2 + aλ + b)][e^(λx)] = 0
double root : [(λ - alpha)^2] [e^(λx)] = 0
My question now is that if I were to do the same thing with the following equation would it follow that
y''' + ay'' + by' + cy +d = 0
Characteristic equation λ^3 + aλ^2 + bλ + c
then making the same substitution ( λ^3 + aλ^2 + bλ + c ) e^(λx)
NOW HERE IS MY QUESTION: Can I write the above as
[(λ - alpha)^3][e^(λx)]
or do I have to write it a different way. Thank you very much.. I know this doesn't really have much to do with the solving of differential equations but it would be very helpful if I knew that that was correct for my assignment.
Thanks.