# calculus-differential equation

posted by on .

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=?

• calculus-differential equation - ,

-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 - ,

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]

• calculus-differential equation - ,

I forgot a - sign - use his :)

• calculus-differential equation - ,

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

• calculus-differential equation - ,

Oh - actually we agree exactly