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Math Help - Differential Equation

  1. #1
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    Differential Equation

    I was wondering if someone could help me with where I am going wrong on this question. I will post the work I have completed, which gave me the wrong solution.
    The Question is:
    du/dt= e^(1t-3.2u) u(0)=2.6

    Find the function u(t) from the differential equation and the initial conditions...

    u(t)=__________

    This is what I have done:

    du/dt = e^t/e^(-3.2u)
    e^(-3.2u) du = e^t dt
    integrate both sides
    -1/3.2 e^(-3.2u) = e^t + c, where c is constant
    e^(-3.2u) = -3.2 e^t + c
    -3.2 u = ln(-3.2 e^t +c)
    u = -1/3.2 ln(-3.2 e^t + c)

    Solving for c by using the initial condition
    u(0) = -1/3.2 ln(-3.2 + c) = 2.6
    Solve for c:
    ln(-3.2+c) = -8.32
    c = e^-8.32 + 3.2
    c = 3.2002

    u(t) = -1/3.2 ln (-3.2 e^t + 3.2002)

    But this is not correct
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  2. #2
    Behold, the power of SARDINES!
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    I was wondering if someone could help me with where I am going wrong on this question. I will post the work I have completed, which gave me the wrong solution.
    The Question is:
    du/dt= e^(1t-3.2u) u(0)=2.6
    \frac{du}{dt}=e^{1t-3.2u}=e^{1t} \cdot e^{-3.2u}

    Seperating the equation we get

    e^{3.2u}du=e^tdt if we integrate both sides

    \int e^{3.2u}du=\int e^tdt \iff \frac{1}{3.2}e^{3.2u}=e^t+C

    You should be able to finish from here
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  3. #3
    Super Member wingless's Avatar
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    Quote Originally Posted by BKennedy View Post
    I was wondering if someone could help me with where I am going wrong on this question. I will post the work I have completed, which gave me the wrong solution.
    The Question is:
    du/dt= e^(1t-3.2u) u(0)=2.6

    Find the function u(t) from the differential equation and the initial conditions...

    u(t)=__________

    This is what I have done:

    du/dt = e^t/e^(-3.2u)
    e^(-3.2u) du = e^t dt
    integrate both sides
    -1/3.2 e^(-3.2u) = e^t + c, where c is constant
    e^(-3.2u) = -3.2 e^t + c
    -3.2 u = ln(-3.2 e^t +c)
    u = -1/3.2 ln(-3.2 e^t + c)

    Solving for c by using the initial condition
    u(0) = -1/3.2 ln(-3.2 + c) = 2.6
    Solve for c:
    ln(-3.2+c) = -8.32
    c = e^-8.32 + 3.2
    c = 3.2002

    u(t) = -1/3.2 ln (-3.2 e^t + 3.2002)

    But this is not correct

    \frac{du}{dt} = e^{t-3.2u}

    \frac{du}{dt} = \frac{e^t}{e^{3.2u}}
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