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Math Help - Find General Solution?

  1. #1
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    Find General Solution?

    <br />
y - x\frac{dy}{dx} = \sqrt{1 + (\frac{dy}{dx})^2}(x^2 + y^2)<br />

    I Have Tried It At my level best but was unable to solve it!
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  2. #2
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    y - x\,\frac{dy}{dx} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,(x^2 + y^2)

    \left(y - x\,\frac{dy}{dx}\right)^2 = \left[1 + \left(\frac{dy}{dx}\right)^2\right](x^2 + y^2)^2

    y^2 - 2xy\,\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 = \left[1 + \left(\frac{dy}{dx}\right)^2\right](x^4 + 2x^2y^2 + y^4)

    y^2 - 2xy\,\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 = x^4 + 2x^2y^2 + y^4 + (x^4 + 2x^2y^2 + y^4)\left(\frac{dy}{dx}\right)^2

    (x^4 + 2x^2y^2 + y^4 - 1)\left(\frac{dy}{dx}\right)^2 + 2xy\,\frac{dy}{dx} + x^4 + 2x^2y^2 + y^4 - y^2 = 0.


    This is now a Quadratic in \frac{dy}{dx}, so solve using the Quadratic Formula.
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  3. #3
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    the solution of dy/sx dont seems to be integrate able at all.

    i dont think this is gonna work!

    btw these question are from section where we have to use substitutions like

    x = r.cost & y = r.sint
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  4. #4
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    Just use the Quadratic formula. It's an equation of the form

    aX^2 + bX + c = 0.

    Here X = \frac{dy}{dx}, a = x^4 + 2x^2y^2 + y^4 -1, b = 2xy, c = x^4 + 2x^2y^2 + y^4 - y^2.


    So the solutions are

    \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.


    It's ok to keep your solution in terms of x and y. You don't need to do any integration at all.
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  5. #5
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    Quote Originally Posted by raj007 View Post
    the solution of dy/sx dont seems to be integrate able at all.

    i dont think this is gonna work!

    btw these question are from section where we have to use substitutions like

    x = r.cost & y = r.sint
    Then why not make that substitution. In the end I got

    r_t^2+r^2-1 = 0.
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  6. #6
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    can u show me how u got

    r_t^2+r^2-1 = 0
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