Question

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

Answer 1

do you mean dy/dt=t(y^2-4)?? sorry i cant understand

Answer 2

ya sorry

Answer 3

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

Answer 4

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

Answer 5

?

Answer 6

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.

Answer 7

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

Answer 8

i dont understand how you get this part 1/4 ln(2−y/2+y)

Answer 9

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