Solving an ODE with variable coefficients

Click For Summary
The discussion revolves around solving the ordinary differential equation (ODE) y'' = c2 / (x^2 + c1*x) * y, where c1 and c2 are constants. The method of Frobenius and regular power series were initially applied, but the results yielded an indefinite infinite series. A solution using hypergeometric functions was suggested, with the power series expansion in terms of 1/x rather than x. The coefficients for the series expansion were derived, revealing a pattern that complicates correlation with recognizable series. The clarification on the equation's structure indicates that c2 is divided by (x^2 + c1*x) and multiplied by y.
c0der
Messages
52
Reaction score
0

Homework Statement


Solve the following:
[/B]
y'' = c2 / (x2 + c1*x) * y
c1, c2 are constants, x is variable

Homework Equations


As above

The Attempt at a Solution


I have used the method of Frobenius and regular power series and obtained an infinite series on top of an infinite series, which is indefinite. What are the normal methods to solve that ODE? It's not quite in the Frobenius or Bessel forms
 
Physics news on Phys.org
I used a regular power series and there was no problem. Just multiply the equation by x^2+c_1 x to get x^2 y''+c_1 x y''-c_2 y=0.
The regular series gives only one of the solutions.So you should also consider a solution of the form y=\sum_{n=0}^\infty a_n x^{n+m}.
 
Last edited:
c0der said:

Homework Statement


Solve the following:
[/B]
y'' = c2 / (x2 + c1*x) * y
c1, c2 are constants, x is variable

Homework Equations


As above

The Attempt at a Solution


I have used the method of Frobenius and regular power series and obtained an infinite series on top of an infinite series, which is indefinite. What are the normal methods to solve that ODE? It's not quite in the Frobenius or Bessel forms

Maple gets a solution in terms of hypergeometric functions with (variable) argument ##-c_1/x##. Thus, the power-series expansion of Maple's solution will be in powers of ##1/x##, not of ##x## itself.
 
Thank you, I have done this however as follows:

Equating coefficients of x0 gives:
a0=0

For x1:

a2 = c2/2c1

Equating coefficients of xm:

am+1 = [ c2 - (m-1)m ] / [ c1m(m+1) ] am for m>=2

Then:

a3 = [ (c2 - 2) / 3!2!c12 ] a1

a4 = [ (c2 - 6)(c2 - 2)c2 / 4!3!c13 ] a1

a5 = [ (c2 - 12)(c2 - 6)(c2 - 2)c2 / 5!4!c14 ] a1

a6 = [ (c2 - 20)(c2 - 12)(c2 - 6)(c2 - 2)c2 / 6!5!c15 ] a1

How can I correlate this with the hypergeometric series or any other recognizable series? The numerator terms in the brackets have a pattern to it 2+4=6 6+6=12 12+8=20 etc but getting the terms in powers of m is difficult
 
c0der said:
y'' = c2 / (x2 + c1*x) * y
Is that (c2 / (x2 + c1*x)) * y or c2 / ((x2 + c1*x) * y)?
 
(c2 / (x2 + c1x)) * y
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K