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calculus-differential equation

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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

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