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goalie2012

  • 2 years ago

consider the differential equation: x^3(x^2-1)^2(x^2+1)y''+(x+1)xy'+y=0. determine whether x=0 is a regular singular point. determine whether x=1 is a regular singular point. are there any regular singular points that are complex numbers? justify conclusions

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  1. goalie2012
    • 2 years ago
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    \[x ^{3}(x^2-1)^2(x^2+1)y''+(x-1)xy'+y=0\]

  2. goalie2012
    • 2 years ago
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    the answers I got for this one were 0, 1, and i, but I'm not sure how to justify them. Are those even right to begin with?

  3. goalie2012
    • 2 years ago
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    if you know this one, please answer, but I'm going to write the other one since I'm having more trouble with it.

  4. oldrin.bataku
    • 2 years ago
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    First, rewrite it in standard form:$$y''+\frac{x(x-1)}{x^3(x^2-1)^2(x^2+1)}y+\frac1{x^3(x^2-1)^2(x^2+1)}y=0$$

  5. oldrin.bataku
    • 2 years ago
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    To see if \(x=0\) is ordinary, we see if the coefficients of \(y'\) and \(y\) are analytic at \(x=0\); as you can see, \(y',y\) are not analytic therefore \(x=0\) is singular.

  6. oldrin.bataku
    • 2 years ago
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    To determine whether it is regular, we consider whether \(xp(x),x^2q(x)\) are analytic at \(x=0\) where \(p(x),q(x)\) are the coefficients of \(y',y\) respectively. Observe that neither \(xp(x),x^2q(x)\) are analytic at \(x=0\) (there is at least one factor of \(x\) remaining in the denominator) thus this is an irregular singular point.

  7. oldrin.bataku
    • 2 years ago
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    for the above I meant the coefficients of \(y',y\) were not singular:$$y''+\underbrace{\frac{x(x-1)}{x^3(x^2-1)^2(x^2+1)}}_{p(x)}y+\underbrace{\frac1{x^3(x^2-1)^2(x^2+1)}}_{q(x)}y=0$$Note that we can simplify \(p(x)\):$$p(x)=\frac{x-1}{x^2(x^2-1)^2(x^2+1)}$$which unfortunately is still not analytic at \(x=0\). Thus it is a singular point. Observe that \(xp(x)\) is also not analytic:$$xp(x)=\frac{x-1}{x(x^2-1)^2(x^2+1)}$$thus \(x=0\) is an irregular singular point.

  8. oldrin.bataku
    • 2 years ago
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    For \(x=1\), consider simplifying \(p(x)\) a little more:$$p(x)=\frac{x-1}{x^2(x^2-1)(x^2-1)(x^2+1)}=\frac1{x^2(x+1)(x^2-1)(x^2+1)}$$yet unfortunately it is still not analytic at \(x=1\)

  9. oldrin.bataku
    • 2 years ago
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    Similarly consider \((x-1)p(x)\):$$(x-1)p(x)=\frac{x-1}{x^2(x+1)(x^2-1)(x^2+1)}=\frac1{x^2(x+1)^2(x^2+1)}$$which is analytic at \(x=1\)! Now, check \((x-1)^2q(x)\):$$(x-1)^2q(x)=\frac{(x-1)^2}{x^3(x^2-1)^2(x^2+1)}=\frac1{x^3(x+1)^2(x^2+1)^2}$$which is also analytic at \(x=1\) hence \(x=1\) is a regular singular point

  10. oldrin.bataku
    • 2 years ago
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    To determine whether there are complex singular points, observe the roots of our denominator of \(p(x)\):$$x^2(x+1)(x^2-1)(x^2+1)=0$$we observe we have singularities at \(x^2+1=0\) i.e. \(x=\pm i\). To determine whether they're regular, consider whether \((x-i)p(x),(x+i)p(x)\) still have the singular point:$$(x-i)p(x)=\frac{x-i}{x^2(x+1)^2(x^2+1)}=\frac1{x^2(x+1)^2(x+i)}\\(x+i)p(x)=\frac{x+i}{x^2(x+1)^2(x^2+1)}=\frac1{x^2(x+1)^2(x-i)}$$nope! Since \(q(x)\)'s denominator has the same root with the same multiplicity you can conclude \((x\pm i)q(x)\) will also be analytic at \(x=\pm i\). Hence \(x=\pm i\) are regular singular points.

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