calculusdifferential equation
posted by Jeff .
Consider the differential equation:
(du/dt)=u^2(t^3t)
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=?

calculusdifferential equation 
Count Iblis
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 
calculusdifferential equation 
Damon
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] 
calculusdifferential equation 
Damon
I forgot a  sign  use his :)

calculusdifferential equation 
Jeff
Thanks so much. I've been struggling with that thing for days!

calculusdifferential equation 
Damon
Oh  actually we agree exactly
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