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Integrating factor problems (differential equation with initial conditions)

  1. Nov 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Find general solution of equation
    (t^3)y' + (4t^2)y = e^-t

    with initial conditions:
    y(-1) = 0 and t<0

    book answer gives y = -(1+t)(e^-t)/t^4 t not = 0


    2. Relevant equations


    3. The attempt at a solution

    (t^3)y' + (4t^2)y = e^-t

    get integrating factor...
    u(t) = e^integ (4t^2)/t^3
    u(t) = t^4

    (t^4)y = integ (e^-t)(t^4)

    (t^4)y = (-t^4)e^-t - (4t^3)e^-t - (12t^2)e^-t - 24te^-t - 24e^-t + c

    y = [ (-t^4)e^-t - (4t^3)e^-t - (12t^2)e^-t - 24te^-t - 24e^-t + c] / (t^4)

    simplifying...

    y = [ e^-t ((-t^4) - (4t^3) - (12t^2) - (24t) - 24 + c) ] / (t^4)

    when i put in inititial condition y(-1) = 0
    i end up with

    0 = (-9e^1) + c
    so, c = 9e
    so solution to general solution is:
    y = [ (-t^4)e^-t - (4t^3)e^-t - (12t^2)e^-t - 24te^-t - 24e^-t + 9e] / (t^4)
    but this is really wrong.

    also, I am confused about the other initial condition.. What does it mean that t must be less than zero with repect to the original equation?? i mean , t already is < 0 because y(-1) = 0

    thanks for any help
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 27, 2009 #2
    You forgot to divide both sides of the equation by the leading term!

    Your ODE in standard form should be

    [tex]y' + (4t^{-1})y = e^{-t}t^{-3}[/tex]

    The good news is that you found the correct integrating factor and you now have an easier integral to evaluate!
     
  4. Nov 27, 2009 #3
    wow that was stupid. yup, when i multiply RHS by t^4 it works out. thanks

    also, what does quetion mean when it says t must be less than zero, when they already specified that t = -1?
     
  5. Nov 27, 2009 #4

    Mark44

    Staff: Mentor

    I haven't worked this all the way through, so can't say for sure, but it could be that your solution might involve a square root in some way. Knowing whether t > 0 or t < 0 would enable you to identify which square root you should choose for your solution.
     
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