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    Is the general solution to this equation:

     x \frac{du}{dx} - \frac{1}{2}y\frac{du}{dy} = 0

    This?
     u(x,y) = ABx^\lambda y^\frac{\lambda}{2} Where A and B are constants of integration.

    If so how would I go about doing this question:
    Determine the u(x,y) which satisfies the boundary condition; u(1,y) = 1+sin (y)
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    anyone?
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    No. The y exponent should be twice the x exponent.

    You can also combine AB into one constant since they are both constant.

    It might help to superpose solutions, i.e.

    u(x,y) = \sum_\lambda (x^2y)^\lambda
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    I actually have no idea what to do. The sin y is confusing me. Do I have have to use  e^{ia} = cos(a) +isin(a) or somehing else?
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    (Original post by flipper_99)
    I actually have no idea what to do. The sin y is confusing me. Do I have have to use  e^{ia} = cos(a) +isin(a) or somehing else?
    Let \displaystyle u(x,y) = \sum_{\lambda = 0}^\infty A_\lambda x^{\frac{\lambda}{2}}y^{\lambda}

    Fix A_\lambda so that if you set x = 1, the sum becomes the expansion of siny.
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    it would be useful to set A_0 = 1 + 1 to get the required first term for the siny expansion
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    Do you mean \displaystyle u(x,y) = \sum_{\lambda = 0}^\infty A_\lambda x^{\lambda}y^{2\lambda}?

    Even so, I still have no idea. Do you mean the taylor expansion for sine? I bet it's something trivial, but I just can't seem to see it. Quite annoying
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    (Original post by flipper_99)
    Do you mean \displaystyle u(x,y) = \sum_{\lambda = 0}^\infty A_\lambda x^{\lambda}y^{2\lambda}?

    Even so, I still have no idea. Do you mean the taylor expansion for sine? I bet it's something trivial, but I just can't seem to see it. Quite annoying
    yeah, sorry. I've edited the previous post to the correct form.

    See http://en.wikipedia.org/wiki/Sine#Series_definitions
 
 
 
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