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Math Help - Help me solve this differential equation

  1. #1
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    Post Help me solve this differential equation

    Hi all,

    this is my first post here. I just registered today. I have a little problem: solving differential equations generally. Our professor at my college sent us some math problems for practice, because we're having a final exam at the end of January. Anyway, the first problem is solving a differential equation and I can't seem to remember how to solve it (step by step). I know the first one should be putting all y's on one side and all x's on the other side of the equation, then we should do the integration and so on.

    My problem looks like this: y' = y/x + (y/x)^2

    How do I solve this problem? I keep getting to a point where it is obvious I'm going the wrong way. If anyone can help me, that person would make me happy, because I've been struggling with this all day now, and still haven't found anything similar that could help me.

    Thanks!
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  2. #2
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    Quote Originally Posted by boris90 View Post
    Hi all,

    this is my first post here. I just registered today. I have a little problem: solving differential equations generally. Our professor at my college sent us some math problems for practice, because we're having a final exam at the end of January. Anyway, the first problem is solving a differential equation and I can't seem to remember how to solve it (step by step). I know the first one should be putting all y's on one side and all x's on the other side of the equation, then we should do the integration and so on.

    My problem looks like this: y' = y/x + (y/x)^2

    How do I solve this problem? I keep getting to a point where it is obvious I'm going the wrong way. If anyone can help me, that person would make me happy, because I've been struggling with this all day now, and still haven't found anything similar that could help me.

    Thanks!
    let \displaystyle u=\frac{y}{x} \implies \frac{du}{dx}=\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2} \iff \frac{dy}{dx}=x\frac{du}{dx}+\frac{y}{x} \iff \frac{dy}{dx}=x\frac{du}{dx}+u

    If you sub this into the ODE you get

    \displaystyle x\frac{du}{dx}+u=u+u^2 \iff x\frac{du}{dx}=u^2

    This is a first order separable ODE.
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  3. #3
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    Make the substitution \displaystyle u = \frac{y}{x} \equiv y = u\,x.

    Then \displaystyle \frac{dy}{dx} = u + x\,\frac{du}{dx}.


    Substituting into your DE \displaystyle \frac{dy}{dx} = \frac{y}{x} + \left(\frac{y}{x}\right)^2 gives

    \displaystyle u + x\,\frac{du}{dx} = u + u^2

    \displaystyle x\,\frac{du}{dx} = u^2

    \displaystyle \frac{du}{dx} = u^2x^{-1}

    \displaystyle u^{-2}\,\frac{du}{dx} = x^{-1}

    \displaystyle \int{u^{-2}\,\frac{du}{dx}\,dx} = \int{x^{-1}\,dx}

    \displaystyle \int{u^{-2}\,du} = \ln{|x|} + C_1

    \displaystyle -u^{-1} + C_2 = \ln{|x|} + C_1

    \displaystyle -\frac{x}{y}= \ln{|x|} + C where \displaystyle C = C_1 - C_2

    \displaystyle -\frac{y}{x} = \frac{1}{\ln{|x|} + C}

    \displaystyle y = -\frac{x}{\ln{|x|} + C}.
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  4. #4
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    How come?

    How come you got \displaystyle x\,\frac{du}{dx} = 1 + u
    How come you got that part? From the previous step, you couldn't get it, 'cause the left side should be equal to u^2 only, not 1 + u.
    Last edited by Jester; December 19th 2010 at 01:40 PM. Reason: fixed latex
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  5. #5
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    !

    Oh, it's nothing. Look like I was looking at something else.
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  6. #6
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    Thanks for help, Prove It! This really helped me. All in all, I think you just need to remember this method of solving problems. I didn't, though. Thanks again!
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