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Math Help - General Solution of Quasilinear Equations

  1. #1
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    General Solution of Quasilinear Equations

    Hi all,

    I understand everything in the following example up until the words "Now, consider..."



    Could somebody help me to understand how we get the \frac{d}{ds}(\frac{x+u}{y}) part?
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  2. #2
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    It's simply the product rule. You are differentiating \frac{x+ u}{y}= (x+ u)\frac{1}{y} with respect to s. Presumably, x and y are functions of s but u, here, is independent of s so we can treat it as a constant. The derivative of (x+ u)\frac{1}{y} is \frac{x+ u}{dx}\frac{1}{y}+ (x+u)\frac{d\frac{1}{y}}{ds}= \frac{dx}{ds}\frac{1}{y}+ (x+ u)\left(-\frac{1}{y^2}\frac{dy}{ds}\right).
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  3. #3
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    Sorry, I should have mentioned that all 3 functions, x, y and u are functions of s here.

    I understand the product rule, but I don't have a clue where the (x+u)/y sprung from.

    My understanding of this problem is that we're trying to find two functions

    \Phi(x,y,u) = constant
    \Psi(x,y,u) = constant

    as the solution of the equation satisfies F(\Phi,\Psi) = 0

    Since \frac{d}{ds}(\frac{x+u}{y}) = 0,

    (x+u)/y is a constant, and so there's \Phi(x,y,u) .

    I just don't understand how we got there.

    In the notes I have, there is alternative way of solving these quasilinear equations, which involves integrating the characteristic equations, then parametrising the initial line and setting s = 0. Does this sound familiar to anyone?
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