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Astrophysics

  • one year ago

@ganeshie8

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  1. Astrophysics
    • one year ago
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    I need help understanding what's going on in this example, \[(4+t^2)\frac{ dy }{ dt }+2ty=4t\] solve the O.D.E then it shows the left side \[(4+t^2)\frac{ dy }{ dt }+2y=\frac{ d }{ dt }[(4+t^2)y]\] \[\frac{ d }{ dt } [(4+t^2)y]=4t\]

  2. Astrophysics
    • one year ago
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    Oops it says \[(4+t^2)\frac{ dy }{ dt }+2ty=\frac{ d }{ dt }[(4+t^2)y]\] forgot the t

  3. Empty
    • one year ago
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    Calculate the derivative on the right hand side of that

  4. ganeshie8
    • one year ago
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    They are simply using the product rule in reverse : \(f'g+fg' = (fg)'\)

  5. ganeshie8
    • one year ago
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    This might be easy to compare against : \(fy' + f'y = (fy)'\)

  6. Astrophysics
    • one year ago
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    OOOOOH

  7. Empty
    • one year ago
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    Yeah, when in doubt, work it out.

  8. Astrophysics
    • one year ago
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    Thanks haha, I think this leads up to the integrating factor, so it's just another of the techniques

  9. ganeshie8
    • one year ago
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    Exactly, that integrating factor business is all about making up the left hand side so that it is ready for using product rule in reverse combined with the fundamental theorem of calc( integral is the inverse of derivative realized up to a constant factor.. )

  10. Astrophysics
    • one year ago
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    I see I see, thanks!

  11. Astrophysics
    • one year ago
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    I've never really used the product rule in reverse, this is nifty

  12. ganeshie8
    • one year ago
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    ** realized up to a constant term

  13. Astrophysics
    • one year ago
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    This is so amazing, omg

  14. Empty
    • one year ago
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    yup it's definitely one of my favorite tricks I wish all ODEs could be solved by integrating factors haha

  15. ganeshie8
    • one year ago
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    Basically we can solve "ANY" differential equation of below form using that integrating factor trick : \[f(t)y' + g(t)y = h(t)\] the solution is given by : \[y(t) = \dfrac{\int (e^{\int (g/f)\, dt }h/f)dt}{e^{\int (g/f)\, dt }}\]

  16. ganeshie8
    • one year ago
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    It is my fav too, and it really becomes powerful in cases where you want to make a DE exact...

  17. Astrophysics
    • one year ago
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    Sweet haha, ok thanks guys, I may have more questions as I read more about ODE xD, as sometimes the book seems to skip steps -.-...

  18. Astrophysics
    • one year ago
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    So if I have \[\frac{ d }{ dt }[(4+t^2)y]=4t \implies d[(4+t^2)y]=4tdt \implies (4+t^2)y=2t^2+C\] so the d cancels out the integral

  19. Astrophysics
    • one year ago
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    I feel I should've known this

  20. Astrophysics
    • one year ago
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    \[(4+t^2)y=\int\limits 4t dt \implies (4+t^2)y=2t^2+C\] to be more clear

  21. Astrophysics
    • one year ago
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    This is so Newtonian!!

  22. ganeshie8
    • one year ago
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    Haha I think it was euler...

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