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Math Help - Hello! Nonlinear Differential Equation

  1. #1
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    Hello! Nonlinear Differential Equation

    Hello, new here. Really need help with this one nonlinear differential equation problem from my Calc IV homework. Any help or hints would be greatly appreciated!

    xy' - y = 1/(x^3)

    Thank you in advance!
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  2. #2
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    Re: Hello! Nonlinear Differential Equation

    ?? That is a linear differential equation! It's integrating factor is easy to find.
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    Re: Hello! Nonlinear Differential Equation

    lolol Then I clearly know less than I thought.
    I tried the integrating factor and got a little bit confused, but that's enough of a hint I guess. Thank you! Will try again & post results.
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  4. #4
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    Re: Hello! Nonlinear Differential Equation

    If it can be rewritten as \displaystyle \begin{align*} \frac{dy}{dx} + P(x)\,y = Q(x) \end{align*}, then you can easily find the integrating factor \displaystyle \begin{align*} e^{\int{P(x)\,dx}} \end{align*}. How can you get your DE into this form?
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  5. #5
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    Re: Hello! Nonlinear Differential Equation

    An integrating factor, for a linear differential equation, f(x)dy/dx+ g(x)y= h(x) is a function, \mu(x), such that multiplying by it makes the left side an "exact derivative". That is, f(x)\mu(x)dy/dx+ g(x)\mu(x)y= d(f(x)\mu(x)y)/dx. Theoretically, all first order differential equations have "integrating factors" but it is only linear equations for which it is easy to find because d(f(x)\mu(x)y)/dx, by the product rule, is f(x)\mu(x) dy/dx+ (f'(x)\mu(x)+ f(x)\mu'(x))y= f(x)\mu(x)dy/dx+ g(x)y so we must have f'(x)\mu(x)+ f(x)\mu'(x)= \mu g(x), a first order, separable, differential equation for \mu. We can separate that as f(x)\mu'(x)= \mu(x)(g(x)- f'(x) so that \frac{d\mu}{\mu}= \frac{g(x)- f'(x)}{f(x)}dx, giving the exponential formula Prove It gave.

    In this case, f(x)= x so f'(x)= 1 and g(x)= -1 so that equation is just \mu+ x\mu'= -\mu which is the same as x\mu'= -2\mu so that we have \frac{d\mu}{\mu}= -\frac{2}{x}dx. That's easy to integrate: ln(\mu)= -2ln(x)+ C. Since we only need one integrating factor, we can take C= 0 and then take the exponential of both sides to get \mu= e^{-2ln(x)}= e^{ln(x^{-2})}= x^{-2} as integrating factor.
    Last edited by HallsofIvy; September 3rd 2012 at 12:31 PM.
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    Re: Hello! Nonlinear Differential Equation

    Thanks, Prove It and HallsofIvy! Those equations really helped. I wish my textbook was as thorough as that post was.

    I worked the integrating factor out and got x^-2, multiplied both sides of the equation and the general solution came out to be -3(x^-2) + Cx.
    Last edited by Nekoteko; September 3rd 2012 at 12:01 PM.
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