## 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.

1. anonymous

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

2. anonymous

ya sorry

3. 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.

4. anonymous

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

5. anonymous

?

6. 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.

7. anonymous

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

8. anonymous

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

9. anonymous

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