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anonymous

  • 4 years ago

Both y = 2e^t and y = e^t are solutions of the differential equation dy/dt = y. Since this equation is autonomous the graph of y=2e^t can be obtained from the graph of y=e^t by horizontal translation. By how much does one have to translate the graph of y=e^t horizontally to obtain the graph of y=2e^t? Does this question make sense?

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  1. TuringTest
    • 4 years ago
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    I'm thinking ln2\[\large e^{t+\ln2}=2e^t\]...makes sense to me at least :D

  2. anonymous
    • 4 years ago
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    so what would the horizontal translation be?

  3. TuringTest
    • 4 years ago
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    ln2 units to the left

  4. TuringTest
    • 4 years ago
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    we are changing\[f(t)\to f(t+\ln2)\]which is a translation of ln2 units left

  5. anonymous
    • 4 years ago
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    do you know what the answer to this is? Find the general solution to (b) dy/dt = t^2y^3 i think i have the right answer but want to be sure.

  6. TuringTest
    • 4 years ago
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    what did you get?

  7. anonymous
    • 4 years ago
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    i got \[y=\pm \sqrt{3/2t^3 + c}\]

  8. TuringTest
    • 4 years ago
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    I think you lost a minus sign, no?

  9. anonymous
    • 4 years ago
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    theres a very good chance .. what should the asnwer look like?

  10. TuringTest
    • 4 years ago
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    \[\int\frac{dy}{y^3}=\int t^2dt\]\[-\frac1{2y^2}=\frac{t^3}3+C\to-\frac3{2t^3+C}=y^2\]\[y=\pm\sqrt{-\frac3{2t^3+C}}\]which would change the domain

  11. anonymous
    • 4 years ago
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    thank you!

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