- #1
danago
Gold Member
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Hey; not much of a homework question, but something i was wanting to find out.
Im still a first year undergraduate and just started on differential equations. We have just finished going over homogeneous 2nd order ODE's of the form:
ay'' + by' + cy = 0
My texbook briefly outlines the solution to any general equation, and from there on simply uses the derived solutions to solve further problems.
The way in which it derives the general solution is by noting that if we let y=e^rx, then the equation becomes:
[tex]
ar^2 e^{rx} + bre^{rx} + ce^{rx} = e^{rx} (ar^2 + br + c) = 0
[/tex]
And since exp(x) is real and positive for all x, the equation only holds true if:
[tex]
ar^2 + br + c = 0
[/tex]
Now that all makes sense to me, but what isn't really clear to me is why we just happen to select y=e^rx? Why not some other function? Is there some proof that shows e^rx as the only function to satisfy the conditions? I realize that e^rx is ideal because of the special property that its derivatives are all scalar multiples of itself, but what reason is there to deny the existence of some other function with similar properties?
Thanks for the help; Even a link pointing me in the right direction is apreciated
Dan.
Im still a first year undergraduate and just started on differential equations. We have just finished going over homogeneous 2nd order ODE's of the form:
ay'' + by' + cy = 0
My texbook briefly outlines the solution to any general equation, and from there on simply uses the derived solutions to solve further problems.
The way in which it derives the general solution is by noting that if we let y=e^rx, then the equation becomes:
[tex]
ar^2 e^{rx} + bre^{rx} + ce^{rx} = e^{rx} (ar^2 + br + c) = 0
[/tex]
And since exp(x) is real and positive for all x, the equation only holds true if:
[tex]
ar^2 + br + c = 0
[/tex]
Now that all makes sense to me, but what isn't really clear to me is why we just happen to select y=e^rx? Why not some other function? Is there some proof that shows e^rx as the only function to satisfy the conditions? I realize that e^rx is ideal because of the special property that its derivatives are all scalar multiples of itself, but what reason is there to deny the existence of some other function with similar properties?
Thanks for the help; Even a link pointing me in the right direction is apreciated
Dan.