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Math Help - system of partial DE's

  1. #1
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    system of partial DE's

    1. The problem statement, all variables and given/known data

    Solve the following system of partial differential equations for u(x,y)

    2. Relevant equations

    du/dy = 2xyu

    du/dx = (y^2 + 5)u

    3. The attempt at a solution

    I am honestly not sure where to start, my lectures and tutorials this week have not been helpful at all. My guess is to take the derivative of the first equation and sub that into the second equation for y and then take the derivative of the second equation to get my final answer. But I am probably completely wrong. Any help or advice would be appreciated!
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  2. #2
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    Re: system of partial DE's

    The notation is \frac{\partial u}{\partial y}=2xyu etc...

    Your solution idea is confusing as stated, what do you mean by take the derivative of the first one? Derivative with respect to what? You mean take \frac{\partial}{\partial x}\frac{\partial u}{\partial y}=\frac{\partial}{\partial y}\frac{\partial u}{\partial x} ?
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  3. #3
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    Re: system of partial DE's

    Actually I tried it and it does noting, lol, it just takes you back to where you started but it would be interesting if someone knew what to do with this.
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  4. #4
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    Re: system of partial DE's

    In \frac{\partial u}{\partial x}= (y^2+ 5)u, y can be treated as a constant: think of it as \frac{du}{dx}= (y^2+ 5)u and separate: \frac{du}{u}= (y^2+ 5)dx where, remember y is a constant. Integrate both sides to get ln|u|= (y^2+ 5)x+ f(y). (Since "y is a constant", the constant of integration may be function of y, f(y).)

    We can solve that for u as u= F(y)e^{(y^2+ 5)x} where F(y)= e^{f(y)}. Since f(y) is an arbitrary function of y, so is F(y).

    With that we can write \frac{\partial u}{\partial y}= F'(y)e^{(y^2+ 5)x}+ F(y)(2xy)e^{(y^2+ 5x)}= 2xyu= 2xye^{(y^2+ 5)x} which, since F cannot depend on x, requires that F(y)= 1 (so that F'(y)= 0).
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  5. #5
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    Re: system of partial DE's

    That simplifies things, I was not treating y as a constant.
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