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Math Help - solve this differential equation

  1. #1
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    solve this differential equation

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  2. #2
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    Quote Originally Posted by szpengchao View Post
    well, it's all about chain rule! looking at u as a function of \xi, \ \eta, and each of \xi and \eta as functions of x, \ y, and

    applying the chain rule we'll get: \frac{\partial u}{\partial x} = \frac{\partial u}{\partial \xi} + \frac{\partial u}{\partial \eta}, and \frac{\partial u}{\partial y}=2 \frac{\partial u}{\partial \xi} + 3\frac{\partial u}{\partial \eta}. now apply the chain rule again, but

    this time for \frac{\partial u}{\partial x} and \frac{\partial u}{\partial y}, to get: \frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial \xi^2}+2\frac{\partial^2 u}{\partial \xi \partial \eta} + \frac{\partial^2 u}{\partial \eta^2}, \ \ \ \frac{\partial^2 u}{\partial y^2}=4\frac{\partial^2 u}{\partial \xi^2}+12\frac{\partial^2 u}{\partial \xi \partial \eta} + 9\frac{\partial^2 u}{\partial \eta^2}, and finally:

    \frac{\partial^2 u}{\partial x \partial y}=2\frac{\partial^2 u}{\partial \xi^2}+5\frac{\partial^2 u}{\partial \xi \partial \eta} + 3\frac{\partial^2 u}{\partial \eta^2}. plugging these three in the equation your problem has given us, will give you:

    \frac{\partial^2 u}{\partial \xi \partial \eta}=-1, which obviously has the general solution u=-\xi \eta + f(\xi) + g(\eta), which, in terms of x, y becomes:

    (*) \ \ \ u(x,y)=-(x+2y)(x+3y)+f(x+2y)+g(x+3y). for the required specific function, first see that:

    \frac{\partial u}{\partial y}=-(5x+12y)+2f'(x+2y)+3g'(x+3y). we'd like to have u(x,0)=0, i.e. f(x)+g(x)=x^2. \ \ \ \ (1)

    we also have this condition that \frac{\partial u}{\partial y}(x,0)=x, which gives us: 2f'(x)+3g'(x)=6x. \ \ \ \ (2)

    solving the system of equations (1) and (2) gives us f(x)=C and g(x)=x^2 - C, where C is a constant.thus

    by (*) the specific solution is u(x,y)=-(x+2y)(x+3y)+(x+3y)^2=y(x+3y). \ \ \ \square
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