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anonymous

  • 5 years ago

du/dt=2+2u+t+tu ?????

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  1. anonymous
    • 5 years ago
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    The right-hand side may be factored as\[2+2u+t+tu=2(1+u)+t(1+u)=(2+t)(1+u)\]

  2. anonymous
    • 5 years ago
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    i can get it to an e function and a t function on each sides but how to i make it a term of a single u ????

  3. anonymous
    • 5 years ago
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    Your equation is then separable:\[\frac{du}{1+u}=(2+t)dt \rightarrow \log (1+u) = 2t+\frac{t^2}{2}+c\]

  4. anonymous
    • 5 years ago
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    of course

  5. anonymous
    • 5 years ago
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    Exponentiate both sides:\[1+u=e^{2t+t^2/2+c}=Ce^{2t+t^2/2} \rightarrow u = Ce^{2t+t^2/2}-1\]

  6. anonymous
    • 5 years ago
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    So, are you going to be a fan :p ?

  7. anonymous
    • 5 years ago
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    i had (1+u).du on the otherside, stupid mistake of not having one over to start with, made it seem a lot more difficult than it really was

  8. anonymous
    • 5 years ago
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    already fanned up last time ;)

  9. anonymous
    • 5 years ago
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    oh! [thumbs up]

  10. anonymous
    • 5 years ago
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    Cool, so you're sorted then...

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spraguer (Moderator)
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