- 190

- 0

y'' - 2y ' + y = 0

y (0) = 0

y ' (0) = 1

Can one assume immediately that

y(0) = c0

and y ' (0) = c1

?

Since these are the first 2 terms in the series?

Thanks!

- Thread starter IniquiTrance
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- 190

- 0

y'' - 2y ' + y = 0

y (0) = 0

y ' (0) = 1

Can one assume immediately that

y(0) = c0

and y ' (0) = c1

?

Since these are the first 2 terms in the series?

Thanks!

[tex]y(x) = \sum_{n = 0}^\infty c_n x^n.[/tex]

Plugging in x = 0 gives y(0) = c

Differentiating

[tex]y'(x) = \sum_{n = 0}^\infty n c_n x^{n - 1}[/tex]

and taking x = 0 gives y'(0) = c

So you can

Science Advisor

Homework Helper

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Then y""= 2y"'- y" so y""(0)= 2y"'(0)- y"= 2- 2= 0. So far, y(x)= 0+ 1x+ (2/2)x

You can continue differentiating y"= 2y'- y to find the higher degree terms.

- 190

- 0

[tex]y(x) = \sum_{n = 0}^\infty c_n x^n.[/tex]

Then as CompuChip says, it follows that whatever this series looks like, c0 will be the only term without an x variable, and is thus equal to y(0).

Likewise, when the series is differentiated, c1 will be the only term with out an x, and thus will equal y '(0).

Is this true?

Science Advisor

Homework Helper

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Yes, of course. That's just the MacLaurin series:

[tex]y(x) = \sum_{n = 0}^\infty c_n x^n.[/tex]

Then as CompuChip says, it follows that whatever this series looks like, c0 will be the only term without an x variable, and is thus equal to y(0).

Likewise, when the series is differentiated, c1 will be the only term with out an x, and thus will equal y '(0).

Is this true?

[tex]\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n[/tex]

The coefficient of x

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