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Math Help - [SOLVED] Differential Equation

  1. #1
    Member ronaldo_07's Avatar
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    [SOLVED] Differential Equation

    Determine the general solution of the exact differential equation
    1− \frac{x}{x^2+y^2}-( \frac{y}{x^2+y^2}) \frac{dy}{dx}=0

    Fix the constant of integration according to the initial condition y(0) = e and write down the explicit form of the solution.

    How do I solve this question?
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  2. #2
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    Quote Originally Posted by ronaldo_07 View Post
    Determine the general solution of the exact differential equation
    1− \frac{x}{x^2+y^2}-( \frac{y}{x^2+y^2}) \frac{dy}{dx}=0

    Fix the constant of integration according to the initial condition y(0) = e and write down the explicit form of the solution.

    How do I solve this question?
    Do you mean

     1 - \frac{x}{x^2+y^2} - \bigg(\frac{y}{x^2+y^2}\bigg)\frac{dy}{dx} = 0 ?
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  3. #3
    Member ronaldo_07's Avatar
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    yes thats the question
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  4. #4
    Member ronaldo_07's Avatar
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    Any help on solving this anyone?
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  5. #5
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    Quote Originally Posted by ronaldo_07 View Post
    Any help on solving this anyone?
    Multiply through by  dx which gives:

     (1-\frac{x}{x^2+y^2})dx - \bigg(\frac{y^2}{x^2+y^2}\bigg) dy = 0

    You now have it in the form:

     f(x,y)dx + g(x,y)dy = 0

    The general solution to an equation of this form is given by:

     \int f(x,y)\partial x + \int \bigg(g(x,y) - \frac{\partial}{\partial y} \int f(x,y)dx \bigg) \partial y = c

    You should get a solution of:

     x - y + x\arctan\bigg(\frac{y^2}{x^2}\bigg) = c
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mush View Post
    Do you mean

     1 - \frac{x}{x^2+y^2} - \bigg(\frac{y}{x^2+y^2}\bigg)\frac{dy}{dx} = 0 ?
    Multiply both sides by x^2+y^2 to get the equation x^2+y^2-x-y\frac{\,dy}{\,dx}=0\implies \frac{\,dy}{\,dx}-y=\left(x^2-x\right)y^{-1}. This is a Bernoulli equation.

    Can you try to take it from here?
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