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Math Help - IVP and Boundary Value Problem

  1. #1
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    IVP and Boundary Value Problem

    Find an interval centered about x=0 for which the given IVP has a unique solution:
    (x-2)y''+3y=x, y(0)=0, y'(0)=1

    what exactly is the question asking?? Do I need to solve the IVP to find the boundaries?
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  2. #2
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    Quote Originally Posted by daskywalker View Post
    Find an interval centered about x=0 for which the given IVP has a unique solution:
    (x-2)y''+3y=x, y(0)=0, y'(0)=1

    what exactly is the question asking?? Do I need to solve the IVP to find the boundaries?
    Well, you could solve it then prove that this solution is unique (not too difficult, I think, since it's a linear equation). Or you could use the existence and uniqueness theorem for linear equations that says that if y''=P(x)+Q(x)y+R(x)y' where P,Q,R are continous functions of x on a<x<b then there is a unique solution to this equation a,b) \rightarrow \mathbb{R}" alt="ya,b) \rightarrow \mathbb{R}" /> with the specified initial conditions on a point.
    Last edited by Jose27; January 28th 2010 at 08:36 PM.
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  3. #3
    MHF Contributor chisigma's Avatar
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    If we suppose that the solution we are searching is analitic in x=0 we can write...

    y(x)= \sum_{n=0}^{\infty} a_{n}\cdot x^{n}

    y^{'}(x)= \sum_{n=1}^{\infty} n\cdot a_{n}\cdot x^{n-1}

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

    From (1) and the 'initial conditions' we derive that is...

    a_{0}=0

    a_{1}=1 (2)

    ... and the DE can be written as...

    3\cdot \sum_{n=2}^{\infty} a_{n}\cdot x^{n} + \sum_{n=2}^{\infty} n\cdot (n-1)\cdot a_{n}\cdot x^{n-1} -2\cdot \sum_{n=2}^{\infty} n\cdot (n-1)\cdot a_{n}\cdot x^{n-2} = -2\cdot x (3)

    If we impose the identity (3) for all n>1 we obtain first...

    a_{2}=0 (4)

    ... and for n>2 ...

    a_{n}= \frac{3\cdot a_{n-2} + (n-1)\cdot (n-2)\cdot a_{n-1}}{2\cdot n\cdot (n-1)} (5)

    The conlusion is that the solution can be written as...

    y(x)= x + \sum_{n=3}^{\infty} \alpha_{n}\cdot (\frac{x}{2})^{n} (6)

    ... and because the sequence \alpha_{n} in (6) is bounded the series in (6) will converge if |x|<2...

    Kind regards

    \chi \sigma
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