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RolyPoly

  • 3 years ago

Differential equation Did I make any mistakes here? \[\frac{dx}{dt} - x^3 = x\]\[\frac{dx}{dt} = x^3 + x\]\[\frac{dx}{x^3+x} = dt\]\[(\frac{1}{x}- \frac{x}{x^2+1})dx = dt\]\[\ln |x| - tan^{-1}x = t + c\]

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  1. zepdrix
    • 3 years ago
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    \[\frac{ dx }{ x^3+1 }=dt\] Hmm why isn't it equal to this? :o Maybe I'm missing something.\[\frac{ dx }{ x^3+x }=dt\]

  2. RolyPoly
    • 3 years ago
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    Sorry, it was a typo!

  3. zepdrix
    • 3 years ago
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    Ah ok i thought so :) imma check the fraction decomp a sec

  4. zepdrix
    • 3 years ago
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    \[\int\limits_{}^{}\frac{ 1 }{ x^2+1 }dx=\tan^{-1}x\] \[\int\limits_{}^{}\frac{ x }{ x^2+1 }dx \neq \tan^{-1}x\] Woops! That's just another natural log I think! :o

  5. RolyPoly
    • 3 years ago
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    Oops! The problem is the partial fraction!!

  6. zepdrix
    • 3 years ago
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    It is? Hmm I got the same thing you did...

  7. RolyPoly
    • 3 years ago
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    Not here. But in my notebook :S

  8. zepdrix
    • 3 years ago
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    oh heh

  9. RolyPoly
    • 3 years ago
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    \[\int\limits_{}^{}\frac{ x }{ x^2+1 }dx = \frac{1}{2}\ln |x^2+1| +C\]

  10. zepdrix
    • 3 years ago
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    yay team c:

  11. RolyPoly
    • 3 years ago
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    Thanks :)

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