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mays

need help to solve this equation e^tdy/dt=y^3

  • one year ago
  • one year ago

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  1. sunnymony
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    e^tdy/dt=y^3 dyv^-3=dt e^-t on integratiion y^-2/2=e^-t/-1 +constant

    • one year ago
  2. mays
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    than you very much

    • one year ago
  3. sunnymony
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    ur welcome

    • one year ago
  4. mays
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    if i need the general solution of the differential equation expressing y explicitly as a function of t and hence find the particular solution that satisfies the initial y(0)=1 do i need to make it y=f(x) then find the constant c then substitute the value of x through f(x) thanks

    • one year ago
  5. mays
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    is it right

    • one year ago
  6. scarydoor
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    Have you written the question correctly? A first order non-linear differential equation? (non-linear?) Wolfram gives: http://www.wolframalpha.com/input/?i=e^t+dy%2Fdt%3Dy^3 A wacky solution... it seems to be interpreting it correctly, as least how I would think of it from what you've written...

    • one year ago
  7. mays
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    yes its true e^t *dy/dt=y^3 (where y is gretethan 0) but i use the wolframalpha but itesnt go right whatdo you thinease

    • one year ago
  8. mays
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    sorry what do you think please

    • one year ago
  9. scarydoor
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    http://www.wolframalpha.com/ Type into the search the problem: e^t *dy/dt=y^3 It will solve it. But it's not a nice looking solution...

    • one year ago
  10. mays
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    the answer i got is y=1/-2e^t+2c

    • one year ago
  11. mays
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    and this without computr

    • one year ago
  12. scarydoor
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    actually the solution the other guy gave is right. But it's written implicitly not explicity, which is why it looks different to what wolfram gives. But still the solution is very ugly. If this is for a class assignment then I think this is either written incorrectly, or it's supposed to look ugly for some reason (maybe it's from a real life problem).

    • one year ago
  13. mays
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    ok thanks lot will try it again

    • one year ago
  14. scarydoor
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    here's the solution that the other guy gave: -1/2 * y^(-2) = -e^(-t) + C. I think that's right. But see how that's written in implicit form. If you solve that for y then I think you should end up with what wolfram alpha got.

    • one year ago
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