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Thread: Question of Differential Equations

  1. #1
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    Question of Differential Equations

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

    what should i do next if this is correct?


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

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

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

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

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

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

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

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

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

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

    $\displaystyle \theta + C_2 = x + C_1$

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

    $\displaystyle y = \sin{(x + C)}$

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

    $\displaystyle y = A\sin{x} + B\cos{x}$, where $\displaystyle A = \cos{C}$ and $\displaystyle B = \sin{C}$.
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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 $\displaystyle y=Ce^{x}$
    $\displaystyle \frac{dy}{dx}=e^x$
    $\displaystyle m=e^x$
    O.T $\displaystyle = \int \frac{-1}{e^x}$
    O.T $\displaystyle = -e^{-x}+C$

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

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

    then

    $\displaystyle y'_{\perp} = - \dfrac{1}{y}$ (you can integrate this giving your result).
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  5. #5
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    thanks
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