Frobenius Method: Finding Solutions for x^2*y"-x*y'+(x^2+1)y=0

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In summary, the conversation is about finding the first and second solutions for the equation x^2*y"-x*y'+(x^2+1)y=0. The general equation for An is given and it is shown that plugging r into this equation gives a solution for y1. However, there is uncertainty about the correctness of the solution and concern about the index value and power of y1. The conversation ends with a request to not post multiple times and a link to a forum for further discussion.
  • #1
jason17349
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Somebody please help, I'm not sure I know what is going on with this.

My problem: find the first solution and use it to find the second solution for

x^2*y"-x*y'+(x^2+1)y=0

assuming y=summation from n=0 to infinity for An*x^n+r

substituting and solving gives me r=1 and a general equation: An=A(n-2)/((n+r)*(n+r-2)+1) for n >= 2

plugging r into my general equation gives An=A(n-2)/((n+1)*(n-1)+1) for n >= 2

plugging n into this I get y=A0*x+(1/4)A0*x^3+(1/64)A0*x^5+(1/2304)A0*x^7... this is y1

now y2=y1*v

I'm not entirely sure what to do after this because I'm unable to reduce y1 to a simple summation which is the only way I've seen this problem done before
 
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  • #2
Maybe you could try to write A_n in product form to see if something more manageable pops out with a nice selection* of A_0.

*You know, the kind of selections people who already know the answer to the problem always do just to mock us mere mortals.
 
  • #3
Using the reduction formula I came up with this solution:

y2=y1*integral(x*(y1)^(-2)*dx)

I don't know how to write math symbols in here so I attached a picture that is easier to understand.

does this seem like the correct solution? I'm also concerned about my answer for y1. The index value and power particularly.
 

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FAQ: Frobenius Method: Finding Solutions for x^2*y"-x*y'+(x^2+1)y=0

1. What is the Frobenius method?

The Frobenius method is a technique used to solve second-order linear differential equations. It involves assuming a solution in the form of a power series and then using the recurrence relation to solve for the coefficients.

2. When is the Frobenius method used?

The Frobenius method is typically used when the differential equation has a regular singular point, meaning that the coefficients of the highest and second-highest order terms are not analytic at that point.

3. What is a regular singular point?

A regular singular point is a point at which the coefficients of the highest and second-highest order terms of a differential equation are not analytic. This means that they cannot be expressed as a power series in the independent variable.

4. What is the difference between a regular singular point and an irregular singular point?

A regular singular point is a point where the coefficients of the highest and second-highest order terms are not analytic. An irregular singular point, on the other hand, is a point where the coefficients of all terms are not analytic, making it more difficult to solve the differential equation using the Frobenius method.

5. Can the Frobenius method be used to solve non-linear differential equations?

No, the Frobenius method is only applicable to linear differential equations. Non-linear differential equations require different methods, such as separation of variables or substitution, to solve.

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