Series Solution to Linear Equations: Finding the Expansion for the Term xy

[WatermelonPig]

Homework Statement

I just can't figure out this one term in the series. For a linear equation, one of the the terms is xy. So I need to find a series expansion of this starting at n = 0.

Homework Equations

The Attempt at a Solution

Assume y = sum(anx^n) n = 0
dy/dx = sum(nanx^(n-1)) n = 1
x(dy/dx) = sum(nanx^n) n = 1
Shifting indexes:
x(dy/dx) = sum((n+1)anx^(n+1)) n = 0
But the solution says:
x(dy/dx) = sum(anx^n) n = 0
So by multiplying the x in does that already shift the index?

 

[LCKurtz]

Homework Statement

I just can't figure out this one term in the series. For a linear equation, one of the the terms is xy. So I need to find a series expansion of this starting at n = 0.

Homework Equations

The Attempt at a Solution

Assume y = sum(anx^n) n = 0
dy/dx = sum(nanx^(n-1)) n = 1
x(dy/dx) = sum(nanx^n) n = 1
Shifting indexes:
x(dy/dx) = sum((n+1)anx^(n+1)) n = 0
But the solution says:
x(dy/dx) = sum(anx^n) n = 0
So by multiplying the x in does that already shift the index?

When you plug a series

y = Σa_nxⁿ

into a DE, as you have noticed, differentiating it reduces the exponent:

y' = Σna_nx^n-1

If you were calculating, for example, y' + y and you add these two series, it is convenient to shift the index on the second series so they are both expressed in terms of xⁿ. That makes it easy to collect terms and figure out any recursion formula.

In your example, multiplying y' by x gives:

xy' = Σna_nxⁿ

so you don't need to shift the index. Circumstances have already given you xⁿ for your general term. The same thing would happen if y'' was multiplied by x². But any other power would require shifting the exponent to get xⁿ as the general term.

 

[WatermelonPig]

Ok, that makes sense. I guess I'lll write out the first few terms from now before deciding on indexing.

 

[LCKurtz]

Ok, that makes sense. I guess I'lll write out the first few terms from now before deciding on indexing.

Just index each sum to give xⁿ. Any extra terms at the beginning of a sum are usually where the arbitrary constants come from.