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Math Help - Hard equation

  1. #1
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    Hard equation

    Hello , I am trying to solve this equation for 2hours but I don't know how to do it...mabye someone can help me? I would be very gratefull Hard equation-untitled.jpg
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  2. #2
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    Re: Hard equation

    Divide through by x, then substitute $\displaystyle \begin{align*} u = y' \end{align*}$. The equation becomes first order linear, and so can be solved with an integrating factor.
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  3. #3
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    Re: Hard equation

    u can also use Runge Kutta method of fourth order
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  4. #4
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    Re: Hard equation

    Yes, though I can't imagine why you would use a complicated numerical method when it is easy to solve for an exact solution.

    What I would do is this: let u= y' so we get the first order equation xu'- 2u= -\frac{2}{x^2}. That is linear so we can separate it, first solving xu'= 2u which in turn can be separated to \frac{du}{u}= 2\frac{dx}{x}. Integrating ln(|u|)= 2ln|x|+ C which is equivalent to u(x)= cx^2.

    To find a "particular solution", use "variation of parameters": seek a solution of the form u(x)= v(x)x^2. Then [tex]u'= v'x^2+ 2xv[tex] and the differential equation becomes xu'- 2u= x^3v'+ 2x^2v- 2x^2v= x^3v'= -\frac{2}{x^2} so v'= -\frac{2}{x^5}.
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  5. #5
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    Re: Hard equation

    yeah thats right but as it is an initial value problem i thought we can apply runge kutta method
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