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Math Help - differential equations

  1. #1
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    differential equations

    I have begun studying differential equations and I think I understand problems involving two variables of x and y, but the first problems they have listed involve e

    dy/dx = e^x+y

    Ok, I would think that you would do something like this:

    dy/dx -e^x = dy/dx e^y

    but really am unsure...
    Please advise.
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  2. #2
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    it is a first-order linear DE

    it is a first-order linear DE
    Attached Thumbnails Attached Thumbnails differential equations-gash.jpg  
    Last edited by CaptainBlack; June 9th 2007 at 09:38 AM.
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  3. #3
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    Quote Originally Posted by confusedagain View Post
    I have begun studying differential equations and I think I understand problems involving two variables of x and y, but the first problems they have listed involve e

    dy/dx = e^x+y

    Ok, I would think that you would do something like this:

    dy/dx -e^x = dy/dx e^y

    but really am unsure...
    Please advise.
    This is an inhomogeneous linear 1-st order ODE. Its general solution is
    the sum of the general solution to the homogeneous equation:

    <br />
\frac{dy}{dx}-y=0<br />

    and a particular integral of:

    <br />
\frac{dy}{dx}-y=e^x<br />

    We solve the homogeneous equation by using a trial solution y=e^{\lambda x}, which when substituted into the equation gives us the
    indicial equation:

    <br />
\lambda - 1=0<br />

    So the general solution of the homogeneous equation is y=Ae^{x}.

    Now as the right hand side of the inhomogeneous equation is in fact a solution of the homogeneous equation tradition dictates that we try y=xe^x out as a particular solution. In this case:

    <br />
\frac{d}{dx}\left(xe^x\right)=e^x + xe^x<br />

    so y=xe^x is indeed a particular integral as expected, so the general solution to the original ODE is:

    <br />
y=Ae^x + xe^x=x^x(A+x)<br />

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by curvature View Post
    it is a first-order linear DE
    Please try to keep the width of attacked images to less than 801 pixels,
    so that the image can be displayed without the need to scroll on smaller
    screens.

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    Please try to keep the width of attacked images to less than 801 pixels,
    so that the image can be displayed without the need to scroll on smaller
    screens.

    RonL
    Thanks for your advice. However I don't know how to control the width. Please tell me how.

    curvature
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  6. #6
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    Quote Originally Posted by curvature View Post
    Thanks for your advice. However I don't know how to control the width. Please tell me how.

    curvature
    Well you can start by not reattaching a larger image after I have resized one for you.

    Copy in image into an image processing tool, any photo editor will do, and
    use the resize tool to set the size to something smaller, then save.

    RonL
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    Well you can start by not reattaching a larger image after I have resized one for you.

    Copy in image into an image processing tool, any photo editor will do, and
    use the resize tool to set the size to something smaller, then save.

    RonL
    The images are directly saved as JPEG format from PPT slides. The images are a little bit wider than expected. However the main page can be seen without scrolling the screen.
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  8. #8
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    the solution

    this image is 800by600
    Attached Thumbnails Attached Thumbnails differential equations-16jun2007.jpg  
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  9. #9
    Forum Admin topsquark's Avatar
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    You could just learn the LaTeX, which isn't all that hard. For example, click on the equation below:
    y = e^{-\int P(x) dx} \left ( \int Q(x) e^{\int P(x) dx} dx + C \right )

    -Dan
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  10. #10
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    Quote Originally Posted by confusedagain View Post
    dy/dx = e^x+y
    I've got a doubt

    Is \frac{dy}{dx}=e^{x+y} or \frac{dy}{dx}=e^x+y ??

    In the first case, would be an equation with separable variables, and the second case like they said before, a linear ODE.
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  11. #11
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    Anyway, I'll solve <br />
\frac{dy}{dx}=e^x+y<br />

    \frac{dy}{dx}=e^x+y\iff{y'-y=e^x}

    Let's find an integrating factor defined by \mu(x)=\exp\left(-\int~dx\right)=e^{-x}

    Now multiplying the equation by this factor yields


    \begin{aligned}<br />
e^{-x}y'-e^{-x}y~&=~1\\<br />
(e^{-x}y)'~&=~1\\<br />
e^{-x}y~&=~\int~dx\\<br />
e^{-x}y~&=~x+c\\<br />
y~&=~xe^x+ce^x<br />
\end{aligned}
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  12. #12
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    clarification of problem

    dy/dx = e^(x+y)
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  13. #13
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    Well, in that case, we have that \frac{dy}{dx}=e^{x+y}\iff\frac{dy}{dx}=e^x\cdot{e^  y}\implies\frac1{e^y}~dy=e^x~dx

    Can you take it from there?
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