anonymous
  • anonymous
Solve the differential equation dy dt = t(y2 − 4) by separation of variables. The integrals involved may require the use of partial fractions.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
do you mean dy/dt=t(y^2-4)?? sorry i cant understand
anonymous
  • anonymous
ya sorry
anonymous
  • anonymous
\[\frac{dy}{dt}=t(y^2-4) \implies \frac{dy}{y^2-4}=tdt \implies \int\limits\frac{dy}{y^2-4}=\int\limits tdt.\] Evaluate the integrals and don't forget to add the constant.

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anonymous
  • anonymous
the end result would be \[y=\pm \sqrt{(Ae^t^2)+4}\]
anonymous
  • anonymous
?
anonymous
  • anonymous
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.
anonymous
  • anonymous
It can also be written as \[\frac{2-y}{2+y}=Ae^{t^2}\] where \(A=ke^2\).
anonymous
  • anonymous
i dont understand how you get this part 1/4 ln(2−y/2+y)
anonymous
  • anonymous
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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