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Thread: Why not find the solution?

  1. #1
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    Why not find the solution?

    Hello everyone, this issue reads:

    Find the orthogonal trajectories of the family of curves  y = C_1e^{-x}

    Well, by definition, the steps are these:

    First solve for C. and C = \displaystyle\frac{y}{e^ {-x}}

    differentiate the equation:
    <br />
 y * (y\prime) + Ce^{-x} = 0 , you get rid of the constant C and finally going to be that, by definition,

     y_2' = - \displaystyle\frac{1}{y'}

    But I have problems to find the answer of the book is

     \displaystyle\frac{y^2}{2} = x + C

    Thank you very much,

    greetings
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  2. #2
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    Remember that if two lines are perpendicular then the product of their gradients is 1.

    The calculation goes like this. If y = Ce^{-x} then y'=-Ce^{-x} = -y. If the curve has gradient -y then the orthogonal curve will have gradient -1/(-y) = 1/y. So we need to solve the differential equation y'=1/y.

    Write it as yy'=1 and integrate to get \tfrac12y^2 = x + C.
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  3. #3
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    Hello! Yes, and I could clarify,
    greetings
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