- #1

asdf1

- 734

- 0

then why can you look at that equation from this view:

total output = response to the input+response to the initial data

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- Thread starter asdf1
- Start date

- #1

asdf1

- 734

- 0

then why can you look at that equation from this view:

total output = response to the input+response to the initial data

- #2

- #3

asdf1

- 734

- 0

i know, but why?

- #4

- #5

asdf1

- 734

- 0

i thought c was a constant? why is that particular one the initial state?

r(x) is the input because usually x is the input, right?

- #6

The integral is in reality a definite one. So:

[tex]y(x) = e^{-h}\int_{x_0}^{x}e^hr(x')dx' + c e^{-h}[/tex]

At [tex]x = x_0[/tex] [tex]e^{-h} = e^{-a.(x_0 - x_0)} = e^0 = 1[/tex] and the integral from [tex]x_0[/tex] to [tex]x_0[/tex] is zero. So,

[tex]y(x_0) = 0 + c.1 = c[/tex]

- #7

asdf1

- 734

- 0

thank you very much! :)

- #8

asdf1

- 734

- 0

by the way, do you mean that a "definite integral" is one that converges?

- #9

No, a definite integral is one that has specified limits, so its result is a number and not a function of the integration variable. As you can see, in my example, I used x' as integration variable and x as one of the limits. Relative to x', x is a constant.asdf1 said:by the way, do you mean that a "definite integral" is one that converges?

- #10

asdf1

- 734

- 0

thanks! :)

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