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Math Help - Question of Differential Equations

  1. #1
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    Question of Differential Equations

    1)Solve y' = \sqrt{1-y^2}
    y=sin(u) dy=cos(u) dy
    y=\int \sqrt{1-y^2}
    y=cos(u)

    what should i do next if this is correct?


    2)Find the orthogonal trajectories for y=Ce^{x}
    \frac{dy}{dx}=e^x
    m=e^x
    O.T = \int \frac{-1}{e^x}
    O.T = -e^{-x}+C

    book's answer is y^2=k-2x

    P.S
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  2. #2
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    \frac{dy}{dx} = \sqrt{1 - y^2}

    \frac{1}{\sqrt{1 - y^2}}\,\frac{dy}{dx} = 1

    \int{\frac{1}{\sqrt{1 - y^2}}\,\frac{dy}{dx}\,dx} = \int{1\,dx}

    \int{\frac{1}{\sqrt{1 - y^2}}\,dy} = x + C_1

    Now make the substitution y = \sin{\theta} so that \theta = \arcsin{y} and dy = \cos{\theta}\,d\theta, then the left hand integral becomes

    \int{\frac{1}{\sqrt{1 - \sin^2{\theta}}}\,\cos{\theta}\,d\theta} = x + C_1

    \int{\frac{\cos{\theta}}{\cos{\theta}}\,d\theta} = x + C_1

    \int{1\,d\theta} = x + C_1

    \theta + C_2 = x + C_1

    \arcsin{y} = x + C where C = C_1 - C_2

    y = \sin{(x + C)}

    y = \sin{x}\cos{C} + \cos{x}\sin{C}

    y = A\sin{x} + B\cos{x}, where A = \cos{C} and B = \sin{C}.
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  3. #3
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    any ideas how the second question should be done?
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  4. #4
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    Quote Originally Posted by Paymemoney View Post


    2)Find the orthogonal trajectories for y=Ce^{x}
    \frac{dy}{dx}=e^x
    m=e^x
    O.T = \int \frac{-1}{e^x}
    O.T = -e^{-x}+C

    book's answer is y^2=k-2x

    P.S
    If y = C e^x then y e^{-x} = C so y'e^{-x} - y e^{-x} = 0 \;\; \text{or}\;\; y' = y. Your orthogonal trajectories satisfies y' \cdot y'_{\perp} = -1

    then

    y'_{\perp} = - \dfrac{1}{y} (you can integrate this giving your result).
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    thanks
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