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Thread: But what *is* this "solution"?

  1. #1
    Oct 2011

    But what *is* this "solution" with repect to to the question?

    I'm watching a KhanAcademy video. The question is to solve the exact diff. eq.
    $\displaystyle y \cos x + 2xe^y + (\sin x +x^2e^y-1)y^\prime=0$

    All of the steps are (obviously) fine and he comes up with
    $\displaystyle \psi(x,y)=y \sin x + x^2 e^y - y + C$

    which he then uses to say the solution is
    $\displaystyle y \sin x + x^2 e^y - y = C$

    I understand those steps, too (well, I think I do). But what is that solution? Is that y? The solution to a diff. eq. should be a function, y, whic, when you take its derivatives, satisfies the original equation.

    So what was $\displaystyle \psi$? Is it just a $\displaystyle y_{general}(x,y)$ which solved the original? Can I say that [tex]\psi(x,y)=y_g(x,y)[\tex]? Note that if I add the partial of $\displaystyle \psi$ wrt x to the partial wrt y times y', I get the original question back.

    I'm not really sure how to ask my question. Does it make sense? Think of this question like a beginning algebra student asking what it means that $\displaystyle x=5$ is the solution to $\displaystyle 2x-10=0$. You answer that question by telling him to plug 5 back into the original equation.)
    Last edited by mathDad; May 11th 2014 at 06:25 AM.
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  2. #2
    MHF Contributor
    Nov 2013

    Re: But what *is* this "solution"?

    This is $y(x)$ expressed in an implicit form because a closed form solution doesn't exist. For a given $C$ and for all $x$ in the domain of $y(x)$ you can solve that equation, somehow, to obtain the value of $y$ corresponding to that $x$ i.e. $y(x)$

    Look at this

    It makes it a bit more clear what $\Psi(x,y)$ is all about.
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