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March 26, 2015

March 26, 2015

Posted by **Jeff** on Tuesday, February 5, 2008 at 5:19pm.

(du/dt)=-u^2(t^3-t)

a) Find the general solution to the above differential equation. (Write the answer in a form such that its numerator is 1 and its integration constant is C).

u=?

b) Find the particular solution of the above differential equation that satisfies the condition u=4 at t=0.

u=?

- calculus-differential equation -
**Count Iblis**, Tuesday, February 5, 2008 at 5:25pm-du/u^2 = (t^3 - t) dt --->

1/u = 1/4 t^4 - 1/2 t^2 + c ---->

u = 1/[1/4 t^4 - 1/2 t^2 + c]

u=4 at t=0 ---> c = 1/4

- calculus-differential equation -
**Damon**, Tuesday, February 5, 2008 at 5:33pmdu/u^2 = -(t^3 -t ) dt

1/u = (1/4)t^4 - (1/2)t^2 + constant

1/u = -t^2 (.5 -.25 t^2) + constant

u = -1/[t^2(.5 - .25 t^2) + C]

4 = -1/C so C = -4

u = -1/[t^2(.5 - .25 t^2) - 4]

- calculus-differential equation -
**Damon**, Tuesday, February 5, 2008 at 5:36pmI forgot a - sign - use his :)

- calculus-differential equation -
**Jeff**, Tuesday, February 5, 2008 at 6:13pmThanks so much. I've been struggling with that thing for days!

- calculus-differential equation -
**Damon**, Tuesday, February 5, 2008 at 8:10pmOh - actually we agree exactly

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