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Math Help - Integration help, how do you do this?

  1. #1
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    Integration help, how do you do this?



    Why does C3(t) = that?
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    So hard to read ><

    Is it \displaystyle C_3'(t) + 9C_3(t) = e^{-t}?

    If so, it's first order linear, so use the Integrating Factor method.
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    Quote Originally Posted by Prove It View Post
    So hard to read ><

    Is it \displaystyle C_3'(t) + 9C_3(t) = e^{-t}?

    If so, it's first order linear, so use the Integrating Factor method.
    Yes it is, whats the integrating factor method?
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    I find it hard to believe that you've been given a first order linear ODE without knowing what the Integrating Factor method is.

    Do you at least know about Separation of Variables?
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    Quote Originally Posted by Prove It View Post
    I find it hard to believe that you've been given a first order linear ODE without knowing what the Integrating Factor method is.

    Do you at least know about Separation of Variables?
    Yeah
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    Quote Originally Posted by Prove It View Post
    I find it hard to believe that you've been given a first order linear ODE without knowing what the Integrating Factor method is.

    Do you at least know about Separation of Variables?
    I get 1/8Ae^-t
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    OK, then read very carefully, for I will post this only once...

    To understand the Integrating Factor method, you need to have a basic idea of the Product Rule, i.e. if \displaystyle u,v are functions of \displaystyle x, then \displaystyle \frac{d}{dx}[u\,v] = u\,\frac{dv}{dx} + v\,\frac{du}{dx}.


    Now examine a general first-order linear ODE:

    \displaystyle \frac{dy}{dx} + P(x)\,y = Q(x).

    What we aim to do is to multiply both sides of the DE by some function of \displaystyle x, call it \displaystyle I(x), so that the LHS is a Product-Rule expansion, which can then be written as a derivative, and therefore be integrated. BTW \displaystyle I(x) is known as the Integrating Factor (hence the name of the method)

    So multiplying both sides by \displaystyle I(x) gives

    \displaystyle I(x)\,\frac{dy}{dx} + I(x)P(x)\,y = I(x)Q(x).


    The LHS will be a Product Rule expansion if \displaystyle I(x)P(x) = \frac{dI}{dx} (Can you see why?)

    This is a separable ODE, so solving for \displaystyle I we have

    \displaystyle P(x) = \frac{1}{I}\,\frac{dI}{dx}

    \displaystyle \int{P(x)\,dx} = \int{\frac{1}{I}\,\frac{dI}{dx}\,dx}

    \displaystyle \int{P(x)\,dx} = \int{\frac{1}{I}\,dI}

    \displaystyle \int{P(x)\,dx} = \ln{I} (You can omit modulus signs and constants because any scalar multiple of the Integrating Factor will do)

    \displaystyle I = e^{\int{P(x)\,dx}}.


    So, if we multiply both sides of a first-order linear ODE by \displaystyle e^{\int{P(x)\,dx}}, we create a Product Rule expansion on the LHS which we can write as a single derivative and integrate both sides. Watch what happens...

    \displaystyle e^{\int{P(x)\,dx}}\,\frac{dy}{dx} + e^{\int{P(x)\,dx}}\,P(x)\,y = e^{\int{P(x)\,dx}}\,Q(x)

    \displaystyle \frac{d}{dx}(e^{\int{P(x)\,dx}}\,y) = e^{\int{P(x)\,dx}}\,Q(x)

    \displaystyle e^{\int{P(x)\,dx}}\,y = \int{e^{\int{P(x)\,dx}}\,Q(x)\,dx}

    \displaystyle y = e^{-\int{P(x)\,dx}}\int{e^{\int{P(x)\,dx}}\,Q(x)\,dx}

    which solves the ODE.


    Looks complicated in theory, in practice, extremely easy. Look at your problem...

    \displaystyle \frac{dC}{dt} + 9C = e^{-t}.

    Here your \displaystyle P(x) = 9, so the Integrating Factor is \displaystyle e^{\int{9\,dt}} = e^{9t}.

    Multiplying both sides by the Integrating Factor gives

    \displaystyle e^{9t}\,\frac{dC}{dt} + 9e^{9t}C = e^{8t}

    \displaystyle \frac{d}{dt}(e^{9t}C) = e^{8t}

    \displaystyle e^{9t}C = \int{e^{8t}\,dt}

    \displaystyle e^{9t}C = \frac{1}{8}e^{8t} + c where \displaystyle c is the Integration Constant

    \displaystyle C = \frac{1}{8}e^{-t} + ce^{-9t}.
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    Quote Originally Posted by Prove It View Post
    Looks complicated in theory, in practice, extremely easy. Look at your problem...

    \displaystyle \frac{dC}{dt} + 9C = e^{-t}.

    Here your \displaystyle P(x) = 9, so the Integrating Factor is \displaystyle e^{\int{9\,dt}} = e^{9t}.

    Multiplying both sides by the Integrating Factor gives

    \displaystyle e^{9t}\,\frac{dC}{dt} + 9e^{9t}C = e^{8t}

    \displaystyle \frac{d}{dt}(e^{9t}C) = e^{8t}

    \displaystyle e^{9t}C = \int{e^{8t}\,dt}

    \displaystyle e^{9t}C = \frac{1}{8}e^{8t} + c where \displaystyle c is the Integration Constant

    \displaystyle C = \frac{1}{8}e^{-t} + ce^{-9t}.
    Thanks i understand how you have got this and have done this for myself. But how does this equal Ae^{-t}
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    It doesn't. You must have misunderstood the question. Ae^{-t} is a solution to that equation only for A= 1/8. Since you have NOT taken a course in differential equations, I suspect you were NOT intended to actually solve that equation. I suspect that the problem was really "Find a value of A such that C_3(t)= Ae^{-t} satisfies the equaton C_3'(t)+ 9C_3(t)= e^{-t}".

    And you can do that just by differentiating the given function and plugging into the equation: if C_3(t)= Ae^{-t} then C_3'(t)= -Ae^{-t} so the equation becomes -Ae^{-t}+ 9Ae^{-t}= e^{-t}. That is the same as 8Ae^{-t}= e^{-t} and since an exponential is never negative, we can divide both sides of the equation by e^{-t} to get 8A= 1 or A= \frac{1}{8}.
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