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Math Help - No idea how to solve this differential equation

  1. #1
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    No idea how to solve this differential equation

    dy/dx + y = x

    hmmmmmmmm
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Kaitosan View Post
    dy/dx + y = x

    hmmmmmmmm
    Integrating factor is e^x

    so multiply the equation by e^x :

    \frac{dy}{dx}\cdot e^x +ye^x=xe^x

    \frac{d}{dx} \left(ye^x\right)=xe^x

    So ye^x=\int xe^x ~dx

    \Rightarrow y=\frac{\int xe^x ~dx}{e^x}

    (use an integration by parts to calculate the integral)
    Last edited by Moo; April 24th 2009 at 08:26 AM.
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    Geez..... I don't know what you did. I've only been doing basic differential equations. I know it's a lot to ask but could you go a bit deeper? That method... integrating factor....? It's all new to me. Also, how can d(ye^x)/dx = e^x(dy/dx +y)?
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    Moo
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    Quote Originally Posted by Kaitosan View Post
    Geez..... I don't know what you did. I've only been doing basic differential equations. I know it's a lot to ask but could you go a bit deeper? That method... integrating factor....? It's all new to me. Also, how can d(ye^x)/dx = e^x(dy/dx +y)?
    Read this : http://www.es.ucsc.edu/~eart111/lecture15.pdf

    It's a method based on the product rule of differentiation


    Okay, let's take a general example :
    y'+p(x)y=q(x)

    The integrating factor is defined as \exp\left(\int p(x) ~dx\right) (it is not necessary to have boundaries)

    Why is it useful ? Because the product rule will help you eliminate y'.

    We indeed have, after multiplying by the integrating factor :
    \exp\left(\int p(x) ~dx\right) \cdot y'+p(x) \exp\left(\int p(x) ~dx\right) y=q(x)\exp\left(\int p(x) ~dx\right) \quad {\color{red}\star}


    But look at the LHS.
    You can see (or you know) that it's the derivative of y\exp\left(\int p(x) ~dx\right)
    ---> apply the product rule (along with the chain rule) to check this :

    \begin{aligned}<br />
\text{derivative} &=y'\exp\left(\int p(x) ~dx\right)+y \left(\exp\left(\int p(x) ~dx\right)\right)' \\<br />
&=y'\exp\left(\int p(x) ~dx\right)+\left(\int p(x) ~dx\right)' \exp\left(\int p(x) ~dx\right) y \\<br />
&=y'\exp\left(\int p(x) ~dx\right)+p(x) \exp\left(\int p(x) ~dx\right)  y \end{aligned}


    So finally, \color{red}\star becomes :
    \left(y \exp\left(\int p(x) ~dx\right)\right)'=p(x) \exp\left(\int p(x) ~dx\right)

    And in order to get y, integrate both sides of the equation. And then divide by \exp\left(\int p(x) ~dx\right)




    In this case, it's "easy", because p(x)=1. And thus the integrating factor is \exp\left(\int dx\right)=e^x


    Am I speakig Chinese to you ?
    Anyway, you can find lots of reference in google, looking for "integrating factor"
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    Bad Chinese perhaps...
    Quote Originally Posted by Moo View Post
    Read this : http://www.es.ucsc.edu/~eart111/lecture15.pdf

    It's a method based on the product rule of differentiation


    Okay, let's take a general example :
    y'+p(x)y=q(x)
    How in the world does this example relate to my problem? All I see is a jumble of letters/functions. Can you please use a better example consisted of only variables "x" and "y" and one that shows how to recognize an equation as one requiring the integrating factor?

    Quote Originally Posted by Moo View Post
    The integrating factor is defined as \exp\left(\int p(x) ~dx\right) (it is not necessary to have boundaries)
    Do you mean e^(integ of x)? If so, is this the only integrating factor? Are there various kinds of integrating factors? If so, how do I determine the appropriate integrating factor?


    Quote Originally Posted by Moo View Post
    <br /> <br />
\frac{dy}{dx}\cdot e^x +ye^x=xe^x<br />
    <br /> <br />
\frac{d}{dx} \left(ye^x\right)=xe^x<br />
    I'm sorry, but I'm not a pro in "calgebra." Please explicitly show the algebraic steps.
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  6. #6
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    Quote Originally Posted by Kaitosan View Post
    Bad Chinese perhaps...


    How in the world does this example relate to my problem? All I see is a jumble of letters/functions. Can you please use a better example consisted of only variables "x" and "y" and one that shows how to recognize an equation as one requiring the integrating factor?



    Do you mean e^(integ of x)? If so, is this the only integrating factor? Are there various kinds of integrating factors? If so, how do I determine the appropriate integrating factor?




    I'm sorry, but I'm not a pro in "calgebra." Please explicitly show the algebraic steps.
    Why don't you tell us what you do know so we can pitch the answers at your level.

    So far you seem to be indicating that you do not know the product rule for derivatives (google for "product rule calculus"), and are not at home with functional notation. If so I doubt we can help you, go back to your instructor and ask for help.

    CB
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  7. #7
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    edit: nvm. I had a mental block throughout this but I think I got it.
    Last edited by Kaitosan; April 24th 2009 at 12:09 PM.
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