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Math Help - [SOLVED] Few ODEs I can't solve

  1. #1
    Member Altair's Avatar
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    [SOLVED] Few ODEs I can't solve

    1st.

    I solved the following question :-

     (\sqrt{x+y} + \sqrt{x-y} )dx - (\sqrt{x+y} - \sqrt{x-y} )dy = 0

    I could solve it till the following step. Got stuck after that.

    \frac{2Vx - v( \sqrt {x+Vx} - \sqrt {x-Vx} ) } { (\sqrt {x+Vx} - \sqrt {x-Vx} ) }= \frac{xdv}{dx}

    Is there any mistake in this? Please solve it further.
    Last edited by Altair; March 4th 2008 at 05:45 PM.
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  2. #2
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    Quote Originally Posted by Altair View Post
    1st.

    I solved the following question :-

     (\sqrt{x+y} + \sqrt{x-y} )dx - (\sqrt{x+y} - \sqrt{x-y} )dy = 0

    I could solve it till the following step. Got stuck after that.

    \frac{2Vx - v( \sqrt {x+Vx} - \sqrt {x-Vx} ) } { (\sqrt {x+Vx} - \sqrt {x-Vx} ) }= \frac{xdv}{dx}

    Is there any mistake in this? Please solve it further.
    \frac{dy}{dx} = \frac{\sqrt{x + y} + \sqrt{x - y}}{\sqrt{x + y} - \sqrt{x - y}} = \frac{\sqrt{x} \, \sqrt{1 + \frac{y}{x}} + \sqrt{x} \, \sqrt{1 - \frac{y}{x}}}{\sqrt{x} \, \sqrt{1 + \frac{y}{x}} - \sqrt{x} \, \sqrt{1 - \frac{y}{x}}} = \frac{\sqrt{1 + \frac{y}{x}} + \sqrt{1 - \frac{y}{x}}}{\sqrt{1 + \frac{y}{x}} - \sqrt{1 - \frac{y}{x}}}


    \Rightarrow v + x \, \frac{dv}{dx} = \frac{\sqrt{1 + v} + \sqrt{1 - v}}{\sqrt{1 + v} - \sqrt{1 - v}} = \frac{(\sqrt{1 + v} + \sqrt{1 - v})}{(\sqrt{1 + v} - \sqrt{1 - v})} \times \frac{(\sqrt{1 + v} + \sqrt{1 - v})}{(\sqrt{1 + v} + \sqrt{1 - v})}


    \frac{1 + \sqrt{1 - v^2}}{v}

    (after expanding and simplifying).


    Therefore x \frac{dv}{dx} = \frac{1 + \sqrt{1 - v^2}}{v} - v = \frac{1 - v^2 + \sqrt{1 - v^2}}{v}.


    Therefore \int \frac{v}{1 - v^2 + \sqrt{1 - v^2}} \, dv = \int \frac{1}{x} \, dx.


    I'm afraid I might have to sleep on it for a while (since it's after midnight right now for me) as I don't see any obvious next step right now.

    I'll make a further reply later if no-one else has stepped in.
    Last edited by mr fantastic; March 5th 2008 at 05:33 AM.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    \frac{dy}{dx} = \frac{\sqrt{x + y} + \sqrt{x - y}}{\sqrt{x + y} - \sqrt{x - y}} = \frac{\sqrt{x} \, \sqrt{1 + \frac{y}{x}} + \sqrt{x} \, \sqrt{1 - \frac{y}{x}}}{\sqrt{x} \, \sqrt{1 + \frac{y}{x}} - \sqrt{x} \, \sqrt{1 - \frac{y}{x}}} = \frac{\sqrt{1 + \frac{y}{x}} + \sqrt{1 - \frac{y}{x}}}{\sqrt{1 + \frac{y}{x}} - \sqrt{1 - \frac{y}{x}}}


    \Rightarrow v + x \, \frac{dv}{dx} = \frac{\sqrt{1 + v} + \sqrt{1 - v}}{\sqrt{1 + v} - \sqrt{1 - v}} = \frac{(\sqrt{1 + v} + \sqrt{1 - v})}{(\sqrt{1 + v} - \sqrt{1 - v})} \times \frac{(\sqrt{1 + v} + \sqrt{1 - v})}{(\sqrt{1 + v} + \sqrt{1 - v})}


    \frac{1 + \sqrt{1 - v^2}}{v}

    (after expanding and simplifying).


    Therefore x \frac{dv}{dx} = \frac{1 + \sqrt{1 - v^2}}{v} - v = \frac{1 - v^2 + \sqrt{1 - v^2}}{v}.


    Therefore \int \frac{v}{1 - v^2 + \sqrt{1 - v^2}} \, dv = \int \frac{1}{x} \, dx.


    I'm afraid I might have to sleep on it for a while (since it's after midnight right now for me) as I don't see any obvious next step right now.

    I'll make a further reply later if no-one else has stepped in.
    The integral on the left is equal to -\ln |\sqrt{1 - v^2} + 1|.

    Proof posted later unless someone steps in first.
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    Quote Originally Posted by mr fantastic View Post
    The integral on the left is equal to -\ln |\sqrt{1 - v^2} + 1|.

    Proof posted later unless someone steps in first.
    Make the substitution u = \sqrt{1 - v^2}.

    Then


    \frac{du}{dv} = \frac{-v}{\sqrt{1 - v^2}} = - \frac{v}{u} \Rightarrow dv = - \frac{u}{v} \, du.


    Then the integral becomes


    \int \frac{v}{u^2 + u} \, \left( - \frac{u}{v} \right) = - \int \frac{u}{u^2 + u} \, du = - \int \frac{1}{u + 1} \, du, \, u \neq 0 \,,


    = - \ln |u + 1| + C = - \ln |\sqrt {1 - v^2} + 1| + C.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    \frac{dy}{dx} = \frac{\sqrt{x + y} + \sqrt{x - y}}{\sqrt{x + y} - \sqrt{x - y}} = \frac{\sqrt{x} \, \sqrt{1 + \frac{y}{x}} + \sqrt{x} \, \sqrt{1 - \frac{y}{x}}}{\sqrt{x} \, \sqrt{1 + \frac{y}{x}} - \sqrt{x} \, \sqrt{1 - \frac{y}{x}}} = \frac{\sqrt{1 + \frac{y}{x}} + \sqrt{1 - \frac{y}{x}}}{\sqrt{1 + \frac{y}{x}} - \sqrt{1 - \frac{y}{x}}}


    \Rightarrow v + x \, \frac{dv}{dx} = \frac{\sqrt{1 + v} + \sqrt{1 - v}}{\sqrt{1 + v} - \sqrt{1 - v}} = \frac{(\sqrt{1 + v} + \sqrt{1 - v})}{(\sqrt{1 + v} - \sqrt{1 - v})} \times \frac{(\sqrt{1 + v} + \sqrt{1 - v})}{(\sqrt{1 + v} + \sqrt{1 - v})}


    \frac{1 + \sqrt{1 - v^2}}{v}

    (after expanding and simplifying).


    Therefore x \frac{dv}{dx} = \frac{1 + \sqrt{1 - v^2}}{v} - v = \frac{1 - v^2 + \sqrt{1 - v^2}}{v}.


    Therefore \int \frac{v}{1 - v^2 + \sqrt{1 - v^2}} \, dv = \int \frac{1}{x} \, dx
    Continuing:

    - \ln |\sqrt{1 - v^2} + 1| = \ln |x| + C


    \Rightarrow \ln \left| \frac{1}{\sqrt{1 - v^2} + 1} \right| = \ln |x| + C


    \Rightarrow \frac{1}{\sqrt{1 - v^2} + 1} = Ax \, where A = e^C


    \Rightarrow \sqrt{1 - v^2} + 1 = \frac{B}{x}, where B = \frac{1}{A}


    \Rightarrow 1 - v^2 = \left( \frac{B}{x} - 1\right)^2.


    Substitute from y = xv and you can probably see what the final couple of lines leading to an answer will be .....
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