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anonymous

  • 4 years ago

Solve the differential equation dy dt = t(y2 − 4) by separation of variables. The integrals involved may require the use of partial fractions.

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  1. anonymous
    • 4 years ago
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    do you mean dy/dt=t(y^2-4)?? sorry i cant understand

  2. anonymous
    • 4 years ago
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    ya sorry

  3. anonymous
    • 4 years ago
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    \[\frac{dy}{dt}=t(y^2-4) \implies \frac{dy}{y^2-4}=tdt \implies \int\limits\frac{dy}{y^2-4}=\int\limits tdt.\] Evaluate the integrals and don't forget to add the constant.

  4. anonymous
    • 4 years ago
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    the end result would be \[y=\pm \sqrt{(Ae^t^2)+4}\]

  5. anonymous
    • 4 years ago
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    ?

  6. anonymous
    • 4 years ago
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    When evaluating the integrals you will should get \[\frac{1}{4}\ln {2-y \over 2+y}=\frac{1}{2}t^2+c \implies \ln {2-y \over 2+y}=2t^2+c \implies {2-y \over 2+y}=ke^{2t^2}\] Solve for y now, if you want.

  7. anonymous
    • 4 years ago
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    It can also be written as \[\frac{2-y}{2+y}=Ae^{t^2}\] where \(A=ke^2\).

  8. anonymous
    • 4 years ago
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    i dont understand how you get this part 1/4 ln(2−y/2+y)

  9. anonymous
    • 4 years ago
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    By evaluating the Left-hand-side integral (\(\large \int {dy \over y^2-4}\)), where you need to use partial fractions.\[\]

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