Differential equations on power series
in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

(1-x²)y''-2xy'+4y=0, given y=∑aᵣxʳ, a power series in x for r∈[0,n], so for r∈[2,n]:




We can expect the power series to be infinite, that is, n→∞.

The DE can now be written:



This simplifies:

4a₀+2a₁x+∑aᵣxʳ(4-r-r²)+∑r(r-1)aᵣxʳ⁻²=0 for r∈[2,n].

This has to be true for all x, so we need to find the constant terms, that is, x⁰ coefficients: 4a₀+2a₂=0 (r-2=0 implies r=2). So a₂=-2a₀.

If we now expand the resulting polynomial a little we get:



Because 4a₀+2a₂=0, this becomes:


We can divide through by x:


Therefore, 2a₁+6a₃=0, and a₃=-⅓a₁.

Progressively, the coefficients can be found. Let A=a₀ and B=a₁ (constants of integration), so a₂=-2A and a₃=-B/3. If we can find out the relation of successive coefficients and hence their relation to A and B, the polynomial can be written in the form y=Ap(x)+Bq(x).

For example, following the above method, we deduce that -2a₂+12a₄=0, a₄=a₂/6=-2A/6=-A/3, and -8a₃+20a₅=0, a₅=2a₃/5=-2B/15.

Therefore we need to find the coefficients of consecutive powers of x as a formula.

The general formula is: aᵣ(4-r-r²)+(r+2)(r+1)aᵣ₊₂=0,


If r=0, a₂=-4a₀/2=-2a₀=-2A, as discovered earlier.

If r=1, a₃=-2a₁/6=-a₁/3=-B/3 as before.



The series in parentheses are both infinite.

by Top Rated User (880k points)

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
86,734 questions
93,245 answers
24,131 users