Series Solutions of Second Order Linear Equations

[Jamin2112]

Homework Statement

(1+x²)y'' - 4xy' + 6y = 0

Homework Equations

I'm going to assume y can be written as [n=0 to ∞] ∑a_nxⁿ

The Attempt at a Solution

y = [n=0 to ∞] ∑a_nxⁿ

----> y' = [n=0 to ∞] ∑(n+1)a_n+1xⁿ
----> y'' = [n=0 to ∞] ∑(n+2)(n+1)a_n+2xⁿ

----> (1+x²)∑(n+2)(n+1)a_n+2xⁿ -4x∑(n+1)a_n+1xⁿ + 6∑(n+1)a_n+1xⁿ = 0.

My problem is that, after multiplying out the functions in front of y,y',y'', everything is summed from n=0 to ∞ but the the exponents on the x's don't match up. For example, one term looks like ∑(n+1)(n+2)aⁿ⁺²xⁿ⁺². I can't just do a transformation j=n+2, making it

[j=-2 to ∞] ∑(j-1)ja_jx^j = (-3)(-2)a_-2x^-2 + (-2)(-1)a_-1x^-1 + 0 + 0 + (1)(2)a₂x² + (2)(3)a₃x³ + ...

because then I have those darned a_-2,a_-1 that screw everything up. In other words, I want everything summed from n=0 to ∞ and only xⁿ being summed. Then I can figure out the recurrence relation, the radius of convergence, etc. Suggestions are appreciated.

NEXT QUESTION:

On another problem, I was able to get to the point of

[n=0 to ∞] ∑xⁿ{(n+2)(n+1)a_n+2 + na_n + 2a_n}=0.

For this equation to be satisfied for all x, the coefficient of each power of x must be zero; hence I conclude that

(n+2)(n+1)a_n+2 + na_n + 2a_n = 0.

I'm now trying to figure out the recurrence relationship.

I write out

2a₀ + 2a₂ = 0 -----> a₂=-a₀
3a₁ + 6a₃ = 0 ------> a₃=(-1/2)a₁
4a₂ + 6a₃ = 0 ------> a₄=-(1/3)a₂=(1/3)a₀
5a₃ + 2a₅ = 0 ------> a₅=-(1/4)a₃=(1/8)a₁
6a₄ + 30a₆ = 0 -----> a₆=-(1/5)a₄=(-1/15)a₀
7a₅ + 42a₇ = 0 ------>a₇=(-1/6)a₅=(-1/48)a₁

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---------> a_n = a_2k = (-1)^k * a₀/((2k-1)(2k-3)) ?

I dunno. Does that seem right? I'm just trying to figure it out in my head. Supposing this is correct, what then do I do with the a_2k+1 stuff--you know, the a₁,a₃,a₅,...

 

[vela]

My problem is that, after multiplying out the functions in front of y,y',y'', everything is summed from n=0 to ∞ but the the exponents on the x's don't match up. For example, one term looks like ∑(n+1)(n+2)aⁿ⁺²xⁿ⁺². I can't just do a transformation j=n+2, making it

[j=-2 to ∞] ∑(j-1)ja_jx^j = (-3)(-2)a_-2x^-2 + (-2)(-1)a_-1x^-1 + 0 + 0 + (1)(2)a₂x² + (2)(3)a₃x³ + ...

because then I have those darned a_-2,a_-1 that screw everything up. In other words, I want everything summed from n=0 to ∞ and only xⁿ being summed. Then I can figure out the recurrence relation, the radius of convergence, etc. Suggestions are appreciated.

The reason you probably can't do it is because it's often impossible. The first few terms typically have to be treated separately as special cases.

On another problem, I was able to get to the point of

[n=0 to ∞] ∑xⁿ{(n+2)(n+1)a_n+2 + na_n + 2a_n}=0.

For this equation to be satisfied for all x, the coefficient of each power of x must be zero; hence I conclude that

(n+2)(n+1)a_n+2 + na_n + 2a_n = 0.

Note you can rewrite this as

a_{n+2} = -\frac{1}{n+1} a_n

So the coefficient of the even powers of x will be proportional to a₀ and the coefficient of the odd powers of x will be proportional to a₁. The series proportional to a₀ is one solution, and the series proportional to a₁ is the other solution.

 

[Jamin2112]

The reason you probably can't do it is because it's often impossible. The first few terms typically have to be treated separately as special cases.

So where do I go from here?

I've been told to:

Solve the given differential equation using a power series about x₀=0. Find the recursion relation and find the first four terms in each of the two linearly independent solutions (unless the series terminates sooner). If possible, find the general term in each solution.

 

[vela]

I think what's confusing you is you're trying to make the limits on the summations match up too early. What you want to do instead is make all the sums contain xⁿ. For instance, the first term gives you

(1+x^2)y'' = \sum_{n=2}^\infty n(n-1)a_n x^{n-2} + \sum_{n=2}^\infty n(n-1)a_n x^n

Now let k=n-2, and rewrite the first sum in terms of k:

(1+x^2)y'' = \sum_{k=0}^\infty (k+2)(k+1)a_{k+2} x^k + \sum_{n=2}^\infty n(n-1)a_n x^n

The first summation will contribute x⁰ and x¹ terms while the second does not, so separate those two out and write the rest as one summation to get:

(1+x^2)y'' = 2a_2 + 6a_3 x + \sum_{n=2}^\infty [(n+2)(n+1)a_{n+2}+n(n-1)a_n] x^n

 

[Jamin2112]

I think what's confusing you is you're trying to make the limits on the summations match up too early. What you want to do instead is make all the sums contain xⁿ. For instance, the first term gives you

(1+x^2)y'' = \sum_{n=2}^\infty n(n-1)a_n x^{n-2} + \sum_{n=2}^\infty n(n-1)a_n x^n

Now let k=n-2, and rewrite the first sum in terms of k:

(1+x^2)y'' = \sum_{k=0}^\infty (k+2)(k+1)a_{k+2} x^k + \sum_{n=2}^\infty n(n-1)a_n x^n

The first summation will contribute x⁰ and x¹ terms while the second does not, so separate those two out and write the rest as one summation to get:

(1+x^2)y'' = 2a_2 + 6a_3 x + \sum_{n=2}^\infty [(n+2)(n+1)a_{n+2}+n(n-1)a_n] x^n

Hmmm... I'll try this and get back to you.