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Math Help - Semi Linear PDE's:2y u_x+u_y=1-5u; 4yu_x-xu_y+2x-4y=0

  1. #1
    Senior Member bugatti79's Avatar
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    Semi Linear PDE's:2y u_x+u_y=1-5u; 4yu_x-xu_y+2x-4y=0

    Folks,

    Could anyone check my work as attached?

    pdf 250 part ii. Find the explicit general solution
    pdf 252. Find the particular solution for the given IC.


    Thanks
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  2. #2
    MHF Contributor
    Jester's Avatar
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    A couple of comments. For your first PDE, you have a 1/2 missing. But on a note, you want to solve the characteristic equations

    \dfrac{dx}{2y} = dy = \dfrac{du}{1-5u}

    Which you pick pairs. You chose

    \dfrac{dx}{2y} = dy

    and

    \dfrac{dx}{2y} = \dfrac{du}{1-5u}.

    Why would you not choose the second pair as

    dy = \dfrac{du}{1-5u}? It so much easier and the y is not needed.

    For your second PDE you have in one of your lines

    u = x + \sqrt{-x^2-2f(-x^2/2-2y^2)}

    Where did this come from?
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    A couple of comments. For your first PDE, you have a 1/2 missing. But on a note, you want to solve the characteristic equations

    \dfrac{dx}{2y} = dy = \dfrac{du}{1-5u}

    Which you pick pairs. You chose

    \dfrac{dx}{2y} = dy

    \dfrac{dx}{2y} = \dfrac{du}{1-5u}.

    Why would you not choose the second pair as

    dy = \dfrac{du}{1-5u}? It so much easier and the y is not needed.
    I guess I overlooked this simpler option. If we use this option, I calculate

    \displaystyle y=-\frac{1}{5}ln(1-5u)+f(x-y^2). rearranging for u I get

    \displaystyle u(x,y)=\frac{1}{5}(-e^{-5(y-f(x-y^2))}+1) which i believe is the same as attachment when the missing 1/2 is factored in.


    Quote Originally Posted by Danny View Post
    For your second PDE you have in one of your lines

    u = x + \sqrt{-x^2-2f(-x^2/2-2y^2)}

    Where did this come from?
    Ok, I believe I have it. Silly mistakes as usual. It should just be -x^2-2(-x^2/2-2y^2) under the square root and I also forgot a constant of integration which would be a function of k. Hence I get the particular solution to be

    u(x,y) = x+2y+x^2+4y^2, this satisfies the original DE and IC's

    Thanks Danny!
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