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Math Help - Clairaut's Equation

  1. #1
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    Clairaut's Equation

    I'm starting on non-linear differential equations and there was a particular example in the book which didn't make much sense to me. It is as follows:

    Solve the differential equation
    \left( x^{2}\; -\; 1 \right)p^{2}\; -\; 2xyp\; +\; y^{2}\; -\; 1\; =\; 0
    They rewrite the equation in this form:
    \left( y\; -\; xp \right)^{2}\; -\; 1\; -\; p^{2}\; =\; 0

    And the say that it "could be broken into two equations, each of Clairaut's form." This assumption confuses me. What is the logic behind it?

    oh yeah! p\; =\; \frac{dy}{dx}
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  2. #2
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    Quote Originally Posted by jameselmore91 View Post
    I'm starting on non-linear differential equations and there was a particular example in the book which didn't make much sense to me. It is as follows:

    Solve the differential equation
    \left( x^{2}\; -\; 1 \right)p^{2}\; -\; 2xyp\; +\; y^{2}\; -\; 1\; =\; 0
    They rewrite the equation in this form:
    \left( y\; -\; xp \right)^{2}\; -\; 1\; -\; p^{2}\; =\; 0

    And the say that it "could be broken into two equations, each of Clairaut's form." This assumption confuses me. What is the logic behind it?

    oh yeah! p\; =\; \frac{dy}{dx}
    Dear ameselmore91,

    \left( y\; -\; xp \right)^{2}\; -\; 1\; -\; p^{2}\; =\; 0

    (y-xp)^2=1+p^2

    (y-xp)=\pm\sqrt{1+p^2}

    y=xp\pm\sqrt{1+p^2}

    This equation is of the Clairaut's form, hence the general solution is,

    y=cx\pm\sqrt{1+c^2}~;~where~c~is~an~arbitary~const  ant.
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