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Solutions to this PDE?

  1. Feb 22, 2013 #1
    I have seen a couple of solutions to this PDE -

    [itex]\frac{\partial x}{\partial u}=\frac{x}{\sqrt{1+y^{2}}}[/itex]

    One is -

    [itex]u=\ln \left | y+\sqrt{1+y^{2}} \right |+f\left ( x \right )[/itex]

    I have no idea how this is arrived at or if it's correct. This is what i want to know.

    The solution i've checked out makes the substitution of [itex]y=\sinh \theta [/itex] giving -

    [itex]u=x\ln \left | \cosh \theta \right | + f\left ( x \right )[/itex]

    which is where i'm a bit stuck as substituting in [itex]\theta =\sinh^{-1} y[/itex] gives a [itex]\cosh \sinh^{-1} y[/itex] term that i don't know how to simpify.

    Any help with both of these would be appreciated
     
  2. jcsd
  3. Feb 22, 2013 #2

    HallsofIvy

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    I am confused by your notation. You have x differentiated with respect to u but on the right both x and y. If you really mean what you have, it is not a "partial" differential equation at all. We are thinking of x as a function of u, with y a fixed parameter. Further that is a separable equation: [itex]dx/du= x\sqrt{1+ y^2}[/itex] gives
    [tex]\frac{dx}{x}= \frac{du}{\sqrt{1+ y^2}}[/tex]
    [tex]ln|x|= \frac{u}{\sqrt{1+ y^2}}+ c[/tex]
    [tex]x= Ce^{\frac{u}{\sqrt{1+ y^2}}}[/tex]
    where [itex]C= e^c[/itex].

    Your original equation treats x as a function of u but your answer gives u as a function of x. Which is intended? if the latter, then
    [tex]\frac{du}{dx}= \frac{\sqrt{1+ y^2}}{x}[/tex]
    [tex]\frac{du}{\sqrt{1+ y^2}}= \frac{dx}{x}[/tex]
    [tex]\frac{u}{\sqrt{1+ y^2}}= ln|x|+ C[/tex]
    [tex]u= (ln|x|+ C)\sqrt{1+y^2}[/tex]
    essentially just solving the previous equation for u.
     
  4. Feb 23, 2013 #3
    Sorry, i've written the dependent and independent variables the wrong way round and got the independent variable wrong. It's been a long week!

    It should be -

    [itex]\frac{\partial u}{\partial y}=\frac{x}{\sqrt{1+y^{2}}}[/itex]

    for a function [itex]u\left ( x,y,z \right )[/itex]

    There will be a constant term in the form [itex]f\left ( x,z \right )[/itex] in addition to whatever the result of the integration yields (which i can't work out).
     
    Last edited: Feb 23, 2013
  5. Feb 26, 2013 #4

    HallsofIvy

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    Since there is no differentation with respect to x, you can treat x as a parameter:
    [tex]\frac{du}{dy}= \frac{x}{\sqrt{1+ y^2}}[/tex]
    is separable
    [tex]du= x\frac{dy}{\sqrt{1+ y^2}}[/tex]
     
  6. Feb 27, 2013 #5
    I've managed to get it sussed. Thanks everyone for the help. Even if it was for me to see that i've written the question down wrong! :smile:
     
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