anonymous
  • anonymous
du/dt=2+2u+t+tu ?????
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
The right-hand side may be factored as\[2+2u+t+tu=2(1+u)+t(1+u)=(2+t)(1+u)\]
anonymous
  • anonymous
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 ????
anonymous
  • anonymous
Your equation is then separable:\[\frac{du}{1+u}=(2+t)dt \rightarrow \log (1+u) = 2t+\frac{t^2}{2}+c\]

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anonymous
  • anonymous
of course
anonymous
  • anonymous
Exponentiate both sides:\[1+u=e^{2t+t^2/2+c}=Ce^{2t+t^2/2} \rightarrow u = Ce^{2t+t^2/2}-1\]
anonymous
  • anonymous
So, are you going to be a fan :p ?
anonymous
  • anonymous
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
anonymous
  • anonymous
already fanned up last time ;)
anonymous
  • anonymous
oh! [thumbs up]
anonymous
  • anonymous
Cool, so you're sorted then...

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