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Homework Help: A Simple PDE

  1. May 18, 2005 #1
    Just need some verification.

    Question 1

    Find the general solutions of the following first order PDE

    [tex]z_x - yz_y = z[/tex]

    Question 2

    Find the general solution of the following first order PDE

    [tex]x^2z_x+y^2z_y = xy[/tex]
     
    Last edited: May 18, 2005
  2. jcsd
  3. May 18, 2005 #2
    Solution to Question 1

    The first thing I did was find the characteristic equations of which I want to combine to form a surface of solutions.

    [tex]\frac{dy}{dx} = -y[/tex]
    [tex]\frac{dz}{dx} = z[/tex]

    Solving these gives me

    [tex]y = c_1e^{-x}[/tex]
    [tex]z = c_2e^x[/tex]

    In proper form...

    [tex]c_1 = ye^x[/tex]
    [tex]c_2 = ze^{-x}[/tex]

    So the general solution is

    [tex]c_2 = g(c_1)[/tex]


    [tex]z = e^xg(ye^x)[/tex]

    Checking this solution is correct, I computed

    [tex]z_x = e^xg(ye^x) + e^xg'(ye^x)ye^x = e^xg(ye^x) + ye^{2x}g'(ye^x) [/tex]

    [tex]z_y = e^xg'(ye^x)e^x = e^{2x}g'(ye^x)[/tex]

    And substitute these into the initial PDE we get

    [tex]z_x - yz_y = e^xg(ye^x) + ye^{2x}g'(ye^x) - y(e^{2x}g'(ye^x))[/tex]
    [tex]\quad = e^xg(ye^x) + ye^{2x}g'(ye^x) - ye^{2x}g'(ye^x)[/tex]
    [tex]\quad = e^xg(ye^x)[/tex]
    [tex]\quad = z[/tex]

    Note that I had earlier defined [itex]z = e^xg(ye^x)[/itex]
     
  4. May 18, 2005 #3

    dextercioby

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    Yes,it looks okay.Can u do the same with the second...?

    Daniel.
     
  5. May 18, 2005 #4
    Characteristic Equations are

    [tex]\frac{dy}{dx} = \frac{y^2}{x^2}[/tex]

    [tex]\frac{dz}{dx} = \frac{y}{x}[/tex]

    Solving these two differential equations...

    (1)
    [tex]\int\frac{dy}{y^2} = \int\frac{dx}{x^2}[/tex]
    [tex]-\frac{1}{y} = -\frac{1}{x} + c_1[/tex]
    [tex]c_1 = \frac{1}{x} - \frac{1}{y}[/tex]

    (2)
    [tex]\int dz = \int\frac{ydx}{x}[/tex]
    [tex]z = y\ln x + c_2[/tex]
    [tex]c_2 = z - y\ln x[/tex]

    Hence the general solution is [itex]c_2 = g\left(\frac{1}{x}-\frac{1}{y}\right)[/itex].

    [tex]z - y\ln x = g\left(\frac{1}{x}-\frac{1}{y}\right)[/tex]

    [tex]z = g\left(\frac{1}{x}-\frac{1}{y}\right) + y\ln x[/tex]

    However when I check this I dont get equality...

    [tex]z_x = g'\left(\frac{1}{x}-\frac{1}{y}\right)\left(-\frac{1}{x^2}\right) + \frac{y}{x}[/tex]

    [tex]z_y = g'\left(\frac{1}{x}-\frac{1}{y}\right)\left(\frac{1}{y^2}\right) + \ln x[/tex]

    And

    [tex]x^2z_x - y^2z_y = -g'\left(\frac{1}{x}-\frac{1}{y}\right)+ \frac{y}{x} - g'\left(\frac{1}{x}-\frac{1}{y}\right) - y^2 \ln x[/tex]
    [tex]\neq g\left(\frac{1}{x}-\frac{1}{y}\right) + y\ln x[/tex]

    And I cant see that I made an error in obtaining [itex]z = g\left(\frac{1}{x}-\frac{1}{y}\right)[/itex]

    Can anyone see where I went wrong?
     
    Last edited: May 18, 2005
  6. May 18, 2005 #5

    saltydog

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    Hello Oxymoron,

    I also get:

    [tex]c_1 = \frac{1}{x} - \frac{1}{y}[/tex]

    However at that point, I am to let:

    [tex]w= \frac{1}{x} - \frac{1}{y}\quad\text{ and }\quad r=y[/tex]

    Letting:

    [tex]V(w,r)\equiv z(x,y)[/tex]

    and substituting into the original equation, I end up with:

    [tex]r^2V_r-\frac{r^2}{wr+1}=0[/tex]

    Solving for V(w,r) I get:

    [tex]V(w,r)=\frac{1}{w}ln(wr+1)+F(w)[/tex]

    substituting back x and y I get:

    [tex]z(x,y)=\frac{xy}{y-x}ln[\frac{y-x}{x}+1]+F[\frac{1}{x}-\frac{1}{y}][/tex]

    That is, I'm not familiar with your use of (2) above. However backsubstitution of this does satisfy the PDE.
     
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