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Homework Help: Integrating factor problems (property of exponents)

  1. Apr 12, 2010 #1
    1. The problem statement, all variables and given/known data

    my integrating factor for the DE ty' + (t+1)y = t is mu(x) = e^integ (1+(1/t))
    = e^(t + ln|t| + c)
    so does this simplify to this...or not?

    = e^t + t + c

    so that DE becomes:

    ((e^t) + t))y = (e^t) + t)

    and then after integrating...

    ((e^t) + t))y = e^t + (1/2)t^2 + c

    is that correct??

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 12, 2010 #2


    Staff: Mentor

    For the integrating factor you don't need the constant.
    Or not. Your integrating factor is et + ln(t). This is NOT EQUAL to et + eln(t). There's a property of exponents you need here that can be used to simplify ab + c. What is it?
  4. Apr 12, 2010 #3
    ty' + (t+1)y = t initial conditions: y(ln(2)) = 1

    mu(x) = e^integ (1+(1/t))
    = e^(t + ln|t|)
    = te^t

    (te^t)y = (te^t)t

    integ ((te^t)y)' = integ(t^2)(e^t)

    (te^t)y = e^t(t^2 - 2t +2 + c)

    ty = t^2 - 2t +2 + c

    y = t-2 + (2/t) + c/t
  5. Apr 12, 2010 #4


    Staff: Mentor

    Your next step was to divide both sides by t to get
    y' + (1 + 1/t)y = 1 (*)

    This is the equation you used to get your integration factor.
    Too many t factors on right side. It should be just te^t.

    The whole reason for find the integration factor is to be able to recognize the left side as the derivative of something. Multiplying (*) on both sides by te^t, we get
    y' te^t + (1 + 1/t)y te^t = te^t

    y' te^t + (y te^t + y e^t)= te^t

    The left side happens to be the derivative of y*te^t, which you can check using the product rule.

    So what we have is
    d/dt[y te^t] = te^t
    [tex]\Rightarrow y te^t = \int t e^t dt[/tex]

    Can you continue from here?
  6. Apr 18, 2010 #5
    I want to clarify why i need to use an integrating factor, to see if im understanding...
    This equation is not separable, that is i cant separate all the t's from the y's. If it were separable it is solvable. correct?
    ty' + (t+1)y = t
    It is not exact either, so the point of using integrating factor is to make the equation exact. And once exact, I can then follow the steps for solving exact equation. correct?

    To be exact, then M_x(x,y) + N_y(x,y) = 0.. That is, derivative of function wrt x + derivative of function wrt y = 0.

    is this correct?
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