- #1

- 143

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If not, is it because of the term's arbritary nature which means we can choose to include/reject it from the determined eigenfunction ?

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- Thread starter JamesGoh
- Start date

- #1

- 143

- 0

If not, is it because of the term's arbritary nature which means we can choose to include/reject it from the determined eigenfunction ?

- #2

- 12

- 0

If, in addition, boundary conditions are given, then the arbitrary constant is set by those conditions.

As a basic example, the equation

[itex]\frac{dx}{dt} + x = 0[/itex]

has a general solution

[itex] x(t) = C e^{-t} [/itex]

but if an initial condition, e.g. [itex] x(0) = 1 [/itex] is given, then the constant is set, and the particular solution for this inital value problem is

[itex] x(t) = e^{-t} [/itex]

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