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anonymous
 one year ago
i'm not able to understand this:
(this is WRT ODE course.)
Linear  When do we say that F(x; y; y1; : : : ; y(n)) = 0 linear?
Write
F(x; y; y1; : : : ; y(n)) = 0
as
L(y)(x) = L(y; y1; : : : ; y(n)) = f(x);
and now L is a linear transformation from
Cn = the space of functions which are at least ntimes
dierentiable
to
F = the space of functions.
If these functions are dened on an open set
R, we say
that the ODE is dened on
. (Why do we take an open set?)
anonymous
 one year ago
i'm not able to understand this: (this is WRT ODE course.) Linear  When do we say that F(x; y; y1; : : : ; y(n)) = 0 linear? Write F(x; y; y1; : : : ; y(n)) = 0 as L(y)(x) = L(y; y1; : : : ; y(n)) = f(x); and now L is a linear transformation from Cn = the space of functions which are at least ntimes dierentiable to F = the space of functions. If these functions are dened on an open set R, we say that the ODE is dened on . (Why do we take an open set?)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0defined* both places sorry about typo

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I believe the "open set" requirement is there for the same reason a function \(f:\mathbb{R}\to\mathbb{R}\) can only be differentiable over an open interval \((a,b)\subset\mathbb{R}\).
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