-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
Consider the differential equation:
(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=?
5 answers
du/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]
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]
I forgot a - sign - use his :)
Thanks so much. I've been struggling with that thing for days!
Oh - actually we agree exactly