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Math Help - Substitution to make an ODE linear.

  1. #1
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    Substitution to make an ODE linear.

    Hello. I need help with a couple of differential equations. I have no idea where to start. Apparently these are Bernoulli differential equations (I'm not even sure), and on Wikipedia it says to solve them
    "A change of variables is made to transform into a linear first-order differential equation."
    How do I change the variables? What's a "linear first-order differential equation" anyway?! How do you solve any of this?

    Any help would be really appreciated.

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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Express both equations in the form y'+p(x)y=q(x)y^n . With the substitution v=y^{1-n} the equations are transformed into linear on v .
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  3. #3
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    Quote Originally Posted by StanceGlanceAdvance View Post
    Hello. I need help with a couple of differential equations. I have no idea where to start. Apparently these are Bernoulli differential equations (I'm not even sure), and on Wikipedia it says to solve them How do I change the variables? What's a "linear first-order differential equation" anyway?! How do you solve any of this?
    [snip]
    Do you not have class notes or access to a textbook?
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  4. #4
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    mr fantastic (damn, that's a good name) sorry about the infraction, won't happen again.

    As for the class notes or a text book, I have neither. The professor is on some convention or something, and the last thing we worked on were definite integrals. However, our assistant has showed us two examples.

    The first was how to solve this by replacing the y with dy/dx and then use integrals to solve it. (which I know how to do now, though it's a rather simple example)

    And the other was using

    I'm afraid that's all I have to work with... Oh, and my test is on Tuesday.

    The education system here is, obviously, a nightmare.
    Last edited by StanceGlanceAdvance; May 27th 2011 at 05:00 AM.
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  5. #5
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    Quote Originally Posted by StanceGlanceAdvance View Post
    The education system here is, obviously, a nightmare.
    Where's "here"?
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  6. #6
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    they are straight-forward bernoulli's equation
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  7. #7
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    "here" is Bosnia. I study Chemistry at the national college in Sarajevo, not that it's of any relevance to my problem.

    Straight-forward? I wouldn't have came to the mathhelpforum if I knew how to solve them, so pls help!
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  8. #8
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    What I recommend is that you go through Chris's DE tutorial's (there's 4 parts) at the top of this forum (they are sticky's). They're really quite informative.
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  9. #9
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    I might be on to something...



    Am I right so far?
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  10. #10
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    Looks good!
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  11. #11
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    Your comment just made my day!

    So now I use this and finish off the damn thing? ( i won't post the actual solution to save space and time)

    The other equation seems to be the same, just might require a bit more of work because of the trig function.(I haven't done it yet)
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  12. #12
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    What I recommend is DO NOT memorize formula's. It's important that you understand the process.
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  13. #13
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    Memorizing formulas is out of my comfort zone too, it's just that I had no theoretical background on this, and I have no idea how the guy(my assistant) got to it. But I'll check the tutorials you recommended.
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  14. #14
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    Suppose you have an equation of the form

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

    If we multiply both sides of the equation by some function of \displaystyle x, we'll call it \displaystyle \mu (x), we get

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

    The aim is to write the left hand side as a single derivative, so we will need find a particular \displaystyle \mu(x) that will make the left hand side a product rule expansion, in other words, of the form \displaystyle \mu\,\frac{dy}{dx} + \frac{d\mu}{dx}\,y. So

    \displaystyle \begin{align*} \frac{d\mu}{dx} &= \mu\,P(x) \\ \frac{1}{\mu}\,\frac{d\mu}{dx} &= P(x) \\ \int{\frac{1}{\mu}\,\frac{d\mu}{dx}\,dx} &= \int{P(x)\,dx} \\ \int{\frac{1}{\mu}\,d\mu} &= \int{P(x)\,dx} \\ \ln{\mu} &= \int{P(x)\,dx} \\ \mu &= e^{\int{P(x)\,dx}}\end{align*}

    So in order to simplify your first order linear DE, you multiply both sides by this integrating factor.

    \displaystyle \begin{align*} \frac{dy}{dx} + P(x)\,y &= Q(x) \\ e^{\int{P(x)\,dx}}\,\frac{dy}{dx} + e^{\int{P(x)\,dx}}\,P(x)\,y &= e^{\int{P(x)\,dx}}\,Q(x) \\ \frac{d}{dx}\left[e^{\int{P(x)\,dx}}\,y\right] &= e^{\int{P(x)\,dx}}\,Q(x) \\ e^{\int{P(x)\,dx}}\,y &= \int{e^{\int{P(x)\,dx}}\,Q(x)\,dx} \\ y &= e^{-\int{P(x)\,dx}}\int{e^{\int{P(x)\,dx}}\,Q(x)\,dx} \end{align*}
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  15. #15
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    This is amazing. Thank you.
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