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Thread: Having problems with a linear 1st order ODE

  1. #1
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    Having problems with a linear 1st order ODE

    I've got a linear 1st order ODE, which I'm trying to solve via integrating factors:

    $\displaystyle \frac{dy}{dx} = 3 - 6x + y - 2xy$


    Book Answer:

    $\displaystyle y = -3 + ce^{x^2 - x}$


    So, from here, this is what I start out with:

    $\displaystyle \frac{dy}{dx} + y(-1 + 2x) = 3 - 6x$

    $\displaystyle P(t) = 2x -1$

    $\displaystyle U = e^{\int 2x - 1 dx} = e^{x^2 - x}$

    $\displaystyle \int \frac{d}{dx}[ye^{x^2 - x}] = \int 3e^{x^2 - x} - 6xe^{x^2 - x} dx$


    But working from here would seem to require integrating $\displaystyle e^{x^2 - x}$, which I didn't think was trivially doable.

    Where am I going wrong?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Having problems with a linear 1st order ODE

    Quote Originally Posted by Lancet View Post
    I've got a linear 1st order ODE, which I'm trying to solve via integrating factors:

    $\displaystyle \frac{dy}{dx} = 3 - 6x + y - 2xy$


    Book Answer:

    $\displaystyle y = -3 + ce^{x^2 - x}$


    So, from here, this is what I start out with:

    $\displaystyle \frac{dy}{dx} + y(-1 + 2x) = 3 - 6x$

    $\displaystyle P(t) = 2x -1$

    $\displaystyle U = e^{\int 2x - 1 dx} = e^{x^2 - x}$

    $\displaystyle \int \frac{d}{dx}[ye^{x^2 - x}] = \int 3e^{x^2 - x} - 6xe^{x^2 - x} dx$


    But working from here would seem to require integrating $\displaystyle e^{x^2 - x}$, which I didn't think was trivially doable.

    Where am I going wrong?
    Put x=x'+2, dx=dx' in the ODE.

    Edit: will not work!
    Last edited by Also sprach Zarathustra; Jun 14th 2011 at 03:02 PM.
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  3. #3
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    Re: Having problems with a linear 1st order ODE

    Quote Originally Posted by Also sprach Zarathustra View Post
    Put x=x'+2, dx=dx' in the ODE.
    I don't understand. I also don't get why this can be done...

    Can you clarify, please?
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  4. #4
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    Re: Having problems with a linear 1st order ODE

    Quote Originally Posted by Lancet View Post
    I've got a linear 1st order ODE, which I'm trying to solve via integrating factors:

    $\displaystyle \frac{dy}{dx} = 3 - 6x + y - 2xy$


    Book Answer:

    $\displaystyle y = -3 + ce^{x^2 - x}$


    So, from here, this is what I start out with:

    $\displaystyle \frac{dy}{dx} + y(-1 + 2x) = 3 - 6x$

    $\displaystyle P(t) = 2x -1$

    $\displaystyle U = e^{\int 2x - 1 dx} = e^{x^2 - x}$

    $\displaystyle \int \frac{d}{dx}[ye^{x^2 - x}] = \int 3e^{x^2 - x} - 6xe^{x^2 - x} dx$


    But working from here would seem to require integrating $\displaystyle e^{x^2 - x}$, which I didn't think was trivially doable.

    Where am I going wrong?
    If you split up your integral you will have problems. Consider your integral (I have factored out a 3)

    $\displaystyle \int 3(1-2x)e^{x^2-x}dx $

    and try the substitution $\displaystyle u = x^2-x$.

    BTW, your initial ODE is separable

    $\displaystyle \dfrac{dy}{dx} = 3 - 6x + y - 2xy = (y+3)(1-2x)$
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Having problems with a linear 1st order ODE

    Quote Originally Posted by Danny View Post
    If you split up your integral you will have problems. Consider your integral (I have factored out a 3)

    $\displaystyle \int 3(1-2x)e^{x^2-x}dx $

    and try the substitution $\displaystyle u = x^2-x$.

    BTW, your initial ODE is separable

    $\displaystyle \dfrac{dy}{dx} = 3 - 6x + y - 2xy = (y+3)(1-2x)$
    How I didn't see that?!
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    Re: Having problems with a linear 1st order ODE

    Quote Originally Posted by Danny View Post

    If you split up your integral you will have problems. Consider your integral (I have factored out a 3)

    Wow, that's a nasty integration example, simple though it is. Thank you for showing that to me!


    Quote Originally Posted by Danny View Post

    BTW, your initial ODE is separable

    <head smack>

    Well, that would have made things considerably easier, had I seen that. Of course, I would have lost out on the integration lesson, so I'm happy with how things turned out.


    Incidentally, is there a trick to spotting what something like this separates out into? Or is it just trial and error?
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    Re: Having problems with a linear 1st order ODE

    Quote Originally Posted by Also sprach Zarathustra View Post
    How I didn't see that?!
    I had the same reaction.
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