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Math Help - First Order DE

  1. #1
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    First Order DE

    Hi,

    I am looking to solve the following First Order differential equation, and am not sure what method to use i.e. seperating variables or Integrating factor.

    The equation is dw/dt = k(a-w)

    The result that the book gives is:

    w = a(1-Be^-kt)

    Can some one please show me how this can be solved in full and what method to use.

    Many thanks,

    Adam.
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  2. #2
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    Quote Originally Posted by aharris1 View Post
    Hi,

    I am looking to solve the following First Order differential equation, and am not sure what method to use i.e. seperating variables or Integrating factor.

    The equation is dw/dt = k(a-w)

    The result that the book gives is:

    w = a(1-Be^-kt)

    Can some one please show me how this can be solved in full and what method to use.

    Many thanks,

    Adam.
    \frac{dw}{dt} = k(a-w) is seperable:


    \int \frac{1}{a - w} \, dw = \int k \, dt.


    Now integrate each side (don't forget the arbitrary constant) and then make w the subject.
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  3. #3
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    Hi Mr Fantastic

    Integrating should give:

    ln(a-w)= kt +C


    taking logs of both sides:

    a-w = e^kt +e^c


    Where do I go from here? How do I deal with the RHS of the equation?

    Many thanks

    Adam
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  4. #4
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    Quote Originally Posted by aharris1 View Post
    Hi Mr Fantastic

    Integrating should give:

    ln(a-w)= kt +C


    taking logs of both sides:

    a-w = e^kt +e^c


    Where do I go from here? How do I deal with the RHS of the equation?

    Many thanks

    Adam
    ln|a-w|=kt+c \iff a-w = e^{kt+c}

    a-w=e^{kt+c} \underbrace{=e^{kt} \cdot e^{c}}_{exp Properties}



    a-w=e^{c}e^{kt}

    isolating w

    w=a-e^{c}e^{kt} now we factor out an "a"

    w=a(1-\frac{e^c}{a}e^{kt})

    now relabeling \frac{e^c}{a} as B

    we get... w=(1-Be^{kt})

    Yeah!!!
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  5. #5
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    First Order DE

    Hi The Empty Set

    I have never seen factoring like that! How does that work?

    I was going to multiply through by -1 to isolate w i.e.

    w-a = -Be^-kt setting e^c = B,a constant

    so:

    w = a-Be^-kt

    Regards,

    Adam
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  6. #6
    Super Member wingless's Avatar
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    Quote Originally Posted by aharris1 View Post
    w = a-Be^-kt
    Now divide both sides by a and see what you get.
    (This is the same as factorizing)
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  7. #7
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    First Order DE

    Hi Wingless

    w/a = (a- Be^-kt)/a

    w/a = 1-Be^-kt

    w = a(1-Be^-kt) .... Right?


    Adam
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  8. #8
    Super Member wingless's Avatar
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    Quote Originally Posted by aharris1 View Post
    Hi Wingless

    w/a = (a- Be^-kt)/a

    w/a = 1-Be^-kt

    w = a(1-Be^-kt) .... Right?
    Well, the thing is,
    \frac{a- B e^{-kt}}{a}

    \frac{a}{a} - \frac{B e^{-kt}}{a}

    1 - \frac{B}{a}. e^{-kt}

    Now, B/a will not be B anymore. We can call it something else, K for example.

    \frac{\omega}{a} = 1 - K. e^{-kt}

    \omega = a(1 - K. e^{-kt})

    I'm not sure where your fault is. You probably forgot to divide both a and Be^(-kt) by a, or you probably forgot to give B/a a new letter.
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  9. #9
    Behold, the power of SARDINES!
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    Quote Originally Posted by aharris1 View Post
    Hi The Empty Set

    I have never seen factoring like that! How does that work?

    I was going to multiply through by -1 to isolate w i.e.

    w-a = -Be^-kt setting e^c = B,a constant

    so:

    w = a-Be^-kt

    Regards,

    Adam
    If you like you can think of it this way "build the fraction" to get the factor you need.

    <br />
w=a-e^{c}e^{-kt} \iff w=a-\frac{a}{a}e^c e^{-kt}<br />

    Now you can factor out the "a"

    w=a-\frac{a}{a}e^c e^{-kt}=a(1-\frac{e^c}{a}e^{-kt})=a(1-Be^{-kt})<br />
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