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Astrophysics

  • one year ago

@ganeshie8

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  1. Astrophysics
    • one year ago
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    How do you find f(x) for a second order dif equation to make it exact, I have the equation \[y''+xy'+y=0\]

  2. Astrophysics
    • one year ago
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    I know if you have the form \[P(x)y''+Q(x)y'+R(x)y=0 \] then the exact equation can be written in the form \[[P(x)y']y'+[f(x)y]'=0\]

  3. Astrophysics
    • one year ago
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    Hmm, I think I can just multiply by an integrating factor

  4. Astrophysics
    • one year ago
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    In this case it's \[\huge e^{\int\limits Q(x)dx}\] right since it's second order

  5. Astrophysics
    • one year ago
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    Haha, nope that's not how it works, I definitely to make it first order

  6. Astrophysics
    • one year ago
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    oooh wait...do I see a product rule

  7. IrishBoy123
    • one year ago
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    yeah! \(y''+xy'+y=0\) \(y''+(xy)^{\prime}=0\) \(y' + xy = C\) then integrating factor which is \(\exp (\int x \, dx) = e^{{x^2 \over 2}}\) so \((y \cdot e^{x^2 \over 2})^{\prime} = C\cdot e^{x^2 \over 2}\) \(y \cdot e^{x^2 \over 2} = C \int \, e^{x^2 \over 2} \, dx\) from wiki: \(\text{... there is no elementary indefinite integral for } \int e^{-x^2}\,dx,\)

  8. Astrophysics
    • one year ago
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    \[(xy)' = xy'+y\]

  9. Astrophysics
    • one year ago
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    Haha yes! I just figured it out

  10. Astrophysics
    • one year ago
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    Thanks @IrishBoy123

  11. Astrophysics
    • one year ago
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    Well this is neat xD

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