Differential Equations through Power Series Expansion

[IniquiTrance]

When solving diff-eq's given initial values, e.g.

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!

 

[CompuChip]

One can assume that
$$y(x) = \sum_{n = 0}^\infty c_n x^n.$$

Plugging in x = 0 gives y(0) = c₀.
Differentiating
$$y'(x) = \sum_{n = 0}^\infty n c_n x^{n - 1}$$
and taking x = 0 gives y'(0) = c₁.

So you can show it, if you assume the series expansion to hold.

 

[HallsofIvy]

You can't "immediately assume" that y(0)= c0 because you haven't yet said what "c0" means! You can immediately assume that y(0)= 0 because you are told that. You are also told that y'(0)= 1 and you can calculate that y"(0)= 2y'(0)- y(0)= 2. Further, since y"= 2y'- y for all x, y"'= 2y"- y' and y"'(0)= 2y"(0)- y'(0)= 2- 1= 1.

Then y""= 2y"'- y" so y""(0)= 2y"'(0)- y"= 2- 2= 0. So far, y(x)= 0+ 1x+ (2/2)x²+ 0x³= x+ x².

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

 

[IniquiTrance]

So if we do assume that a solution will take the form of:

$$y(x) = \sum_{n = 0}^\infty c_n x^n.$$

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 without an x, and thus will equal y '(0).

Is this true?

 

[HallsofIvy]

So if we do assume that a solution will take the form of:

$$y(x) = \sum_{n = 0}^\infty c_n x^n.$$

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 without an x, and thus will equal y '(0).

Is this true?

Yes, of course. That's just the MacLaurin series:
$$\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$$
The coefficient of xⁿ for every n is the n^th derivative divided by n!.