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stottrupbailey

  • 2 years ago

calc II separable equation, please help :)

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  1. stottrupbailey
    • 2 years ago
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    \[\frac{ du }{ dt }=e ^{6u+8t}\] given that u(0)=8. So this is how I set it up \[\int\limits_{}^{}e ^{-6u}du=\int\limits_{}^{}e ^{8t}dt\]

  2. stottrupbailey
    • 2 years ago
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    Is that right so far?

  3. nubeer
    • 2 years ago
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    yes looks right. so far.

  4. stottrupbailey
    • 2 years ago
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    Okay, so this is what I got for my integration: \[\frac{ -e ^{-6u} }{ 6 }=\frac{ e ^{8t} }{ 8 }+c\]

  5. stottrupbailey
    • 2 years ago
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    and then using that u(0)=8, I found that c=-0.125.

  6. stottrupbailey
    • 2 years ago
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    I actually have already worked through this problem, but I got the wrong answer at the end so I'm trying to figure out where I'm going wrong

  7. stottrupbailey
    • 2 years ago
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    so then I end up with \[\frac{ -e ^{-6u} }{ 6 }=\frac{ e ^{8t} }{ 8 }-0.125\]

  8. stottrupbailey
    • 2 years ago
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    so then my final answer was u(t)=\[\frac{ 1 }{ 6 }\ln(\frac{ 3 }{ 4 }e ^{8t}-\frac{ 3 }{ 4 })\]

  9. stottrupbailey
    • 2 years ago
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    but the book says that isn't the right answer

  10. nubeer
    • 2 years ago
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    hmm well i also got c same as u.. well what answer book says?

  11. stottrupbailey
    • 2 years ago
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    well, it's actually webwork, so it just tells me that my answer is wrong, I'm not sure what the right answer is

  12. stottrupbailey
    • 2 years ago
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    I'm thinking something must be wrong with the way I'm doing the algebra to rearrange it in terms of u(t)?

  13. nubeer
    • 2 years ago
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    well i am not sure but the way u solved question looks fine and the answer should be the one u got..

  14. stottrupbailey
    • 2 years ago
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    hmmm. Okay. I'll ask the tutor at 8. Thanks anyway :)

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