Series solution to diffeq, stuck on matching the indices

In summary, a series solution to a differential equation is an infinite sum of terms used to approximate the solution when an exact solution cannot be found. The indices for a series solution are determined by examining the power series representation of the differential equation. Matching the indices involves selecting appropriate coefficients for each term in the series so that it aligns with the determined indices. Common techniques for matching the indices include algebraic manipulation, substitution, and comparing coefficients of like powers. However, there may be challenges in determining convergence, finding appropriate coefficients, and checking for singularities or discontinuities.
  • #1
freshman2013
43
0

Homework Statement



find the series solution to y''+x^2*y'+y=0

Homework Equations


y=summation from n=0 to infinity Cn*x^n


The Attempt at a Solution



y=sum from 0 to inf Cnxn
x^2*y'=sum from 1 to inf nC n xn+1 = sum from 2 to inf (n-1) C n-1 xn = sum from 1 to inf (n-1) C n-1 xn
y''=(n+2)(n+1)Cn+2 xn from 0 to infinity

basically, I can't get all three series to match at starting index n=0 and get them to have the terms x^n
 
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  • #2
It's a little hard to read without Latex, but your y'' doesn't seem right. The y' alone will have terms of the form ##nC_nx^{n-1}## starting at n = 1 (since the constant term vanishes) , so the y'' terms should be ##n(n-1)C_nx^{n-2}## starting at n = 2.

You do not have to fool around with the indices on ##x^2y'##. Keep things as simple as you can, and I think it will match up.
 
  • #3
@brmath my y'' is what you have w/ the indices shifted from 2 to 0. How do you manage not shift indices? Don't you need to combine everything into a single summation term and factor x^n, setting the rest equal to zero? That's what I been doing on every problem so far for finding series to diffeqs.
 
  • #4
freshman2013 said:

Homework Statement



find the series solution to y''+x^2*y'+y=0

Homework Equations


y=summation from n=0 to infinity Cn*x^n


The Attempt at a Solution



y=sum from 0 to inf Cnxn
$$y = \sum_{n=0}^\infty c_n x^n$$
x^2*y'=sum from 1 to inf nC n xn+1 = sum from 2 to inf (n-1) C n-1 xn = sum from 1 to inf (n-1) C n-1 xn
$$x^2 y' = \sum_{n=1}^\infty n c_n x^{n+1} = \sum_{n=2}^\infty (n-1)c_{n-1} x^n = \sum_{n=1}^\infty (n-1)c_{n-1}x^n$$ Looks fine, though that last summation seems unnecessary.

y''=(n+2)(n+1)Cn+2 xn from 0 to infinity
$$y'' = \sum_{n=0}^\infty (n+2)(n+1)c_{n+2} x^n$$ That's fine.

basically, I can't get all three series to match at starting index n=0 and get them to have the terms x^n
You don't have to get them to all start at n=0. You have
$$y'' + x^2 y' + y = \sum_{n=0}^\infty (n+2)(n+1)c_{n+2} x^n + \sum_{n=2}^\infty (n-1)c_{n-1} x^n + \sum_{n=0}^\infty c_n x^n$$ Just separate out the first two terms of the first and last series, e.g.
$$y = c_0 + c_1x + \sum_{n=2}^\infty c_nx^n$$ and then you'll have three summations that all start from n=2 so you can combine them.
 
  • #5
what do I do with the extra C0 and C1X?
 
  • #6
Vela told you that- separate the n= 0 and n= 1 terms from the first and third sums.
$$y'' + x^2 y' + y = \sum_{n=0}^\infty (n+2)(n+1)c_{n+2} x^n + \sum_{n=2}^\infty (n-1)c_{n-1} x^n + \sum_{n=0}^\infty c_n x^n= 0$$

In order that a power series be 0 for all x, we must have each coefficient 0.

If n= 0, [itex](n+2)(n+1)c_{n+2}= 2c_2[/itex] and [itex]c_n= c_0[/itex] so we must have 2c_2+ c_0= 0.

If n= 1, [itex](n+2)(n+1)c_{n+2}= 6c_3[/itex] and [itex]c_n= c_1[/itex] so we must have [itex]6c_3+ c_1= 0[/itex].

For [itex]n\ge 2[/itex] we have all three sums so
[itex](n+2)(n+1)c_{n+2}+ (n- 1)c_{n-1}+ c_n= 0[/itex]

In order to solve for [itex]c_2[/itex], you will need to know [itex]c_0[/itex] and, in order to solve for [itex]c_3[/itex], you will need to know [itex]c_1[/itex]. Those will be the two "constants" for the general solution.
 

1. What is a series solution to a differential equation?

A series solution to a differential equation is an infinite sum of terms that approximates the solution to a given differential equation. It is often used when an exact solution cannot be found using other methods.

2. How do you determine the indices for a series solution to a differential equation?

The indices for a series solution can be determined by examining the power series representation of the given differential equation. The indices correspond to the powers of the independent variable in the power series.

3. What is meant by "matching the indices" in a series solution?

Matching the indices refers to selecting the appropriate coefficients for each term in the series solution so that it aligns with the indices determined from the power series representation of the differential equation. This ensures that the series satisfies the given differential equation.

4. What are some common techniques for matching the indices in a series solution?

Some common techniques for matching the indices include using algebraic manipulation, substitution, and comparing coefficients of like powers of the independent variable. It may also be helpful to use known power series expansions to simplify the process.

5. What are some challenges that may arise when using a series solution to a differential equation?

One challenge is determining the convergence of the series solution, as it may not always converge to the actual solution. Additionally, finding the appropriate coefficients for each term can be a time-consuming process, and may require some trial and error. It is also important to check for any singularities or discontinuities in the solution.

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