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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.
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you mean dy/dt=t(y^24)?? sorry i cant understand

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dy}{dt}=t(y^24) \implies \frac{dy}{y^24}=tdt \implies \int\limits\frac{dy}{y^24}=\int\limits tdt.\] Evaluate the integrals and don't forget to add the constant.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the end result would be \[y=\pm \sqrt{(Ae^t^2)+4}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0When evaluating the integrals you will should get \[\frac{1}{4}\ln {2y \over 2+y}=\frac{1}{2}t^2+c \implies \ln {2y \over 2+y}=2t^2+c \implies {2y \over 2+y}=ke^{2t^2}\] Solve for y now, if you want.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It can also be written as \[\frac{2y}{2+y}=Ae^{t^2}\] where \(A=ke^2\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont understand how you get this part 1/4 ln(2−y/2+y)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0By evaluating the Lefthandside integral (\(\large \int {dy \over y^24}\)), where you need to use partial fractions.\[\]
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