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Thread: First Order Ordinary differential Equation

  1. #1
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    First Order Ordinary differential Equation

    I have the first order Initial Value Problem ODE of

    $\displaystyle y' = 1 + 4y^2$ , $\displaystyle y = \frac{1}{2}*tan(2x+C)$ , $\displaystyle y(0) = 0$

    I calculate that $\displaystyle y' = sec^2(2x+c)$

    $\displaystyle
    sec^2(2x+ C) = 1 + 4(\frac{1}{2}tan(2x+C))^2
    $

    $\displaystyle sec^2(2x+ C) = 1 + 4(\frac{1}{4}tan^2(2x+C))$

    $\displaystyle sec^2(2x+ C) = 1 + tan^2(2x+C))$

    *** trig identity$\displaystyle sec^2(u) = 1 + tan^2(u)$ ***

    y(0) = 0 so

    $\displaystyle sec^2(2x+C) = 1 + tan^2(2x+C))$

    $\displaystyle sec^2(2(0) + C) = 1 + tan^2(2(0) + C)$

    $\displaystyle sec^2(C) = 1 + tan^2(C)$

    $\displaystyle \frac{1}{\frac{sin(C)}{cos(C)}} = 1 + \frac{sin(C)}{cos(C)}$

    $\displaystyle \frac{cos(C)}{sin(C)} = 1 + \frac{sin(C)}{cos(C)}$

    $\displaystyle cos^2(C) = 1 + sin^2(C)$

    $\displaystyle C = n* \pi$

    $\displaystyle cos^2(0) = 1 + sin^2(0)$

    1 = 1

    I don't know if this is proof that y(0) = 0
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  2. #2
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    Quote Originally Posted by Fallen186 View Post
    I have the first order Initial Value Problem ODE of

    $\displaystyle y' = 1 + 4y^2$ , $\displaystyle y = \frac{1}{2}*tan(2x+C)$ , $\displaystyle y(0) = 0$

    I calculate that $\displaystyle y' = sec^2(2x+c)$
    You have this as your LHS.

    now evaluate $\displaystyle 1 + 4y^2$ to get the same.

    $\displaystyle y(0)=0$ is used to solve for C.
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    Life is complex: it has both real and imaginary components.
    Try graphing it
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  4. #4
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    Quote Originally Posted by Fallen186 View Post
    I have the first order Initial Value Problem ODE of

    $\displaystyle y' = 1 + 4y^2$ , $\displaystyle y = \frac{1}{2}*tan(2x+C)$ , $\displaystyle y(0) = 0$

    I calculate that $\displaystyle y' = sec^2(2x+c)$

    $\displaystyle
    sec^2(2x+ C) = 1 + 4(\frac{1}{2}tan(2x+C))^2
    $

    $\displaystyle sec^2(2x+ C) = 1 + 4(\frac{1}{4}tan^2(2x+C))$

    $\displaystyle sec^2(2x+ C) = 1 + tan^2(2x+C))$

    *** trig identity$\displaystyle sec^2(u) = 1 + tan^2(u)$ ***

    y(0) = 0 so

    $\displaystyle sec^2(2x+C) = 1 + tan^2(2x+C))$

    $\displaystyle sec^2(2(0) + C) = 1 + tan^2(2(0) + C)$

    $\displaystyle sec^2(C) = 1 + tan^2(C)$

    $\displaystyle \frac{1}{\frac{sin(C)}{cos(C)}} = 1 + \frac{sin(C)}{cos(C)}$

    $\displaystyle \frac{cos(C)}{sin(C)} = 1 + \frac{sin(C)}{cos(C)}$

    $\displaystyle cos^2(C) = 1 + sin^2(C)$

    $\displaystyle C = n* \pi$

    $\displaystyle cos^2(0) = 1 + sin^2(0)$

    1 = 1

    I don't know if this is proof that y(0) = 0
    $\displaystyle \frac{dy}{dx} = 1 + 4y^2$

    $\displaystyle \frac{dx}{dy} = \frac{1}{1 + 4y^2}$

    $\displaystyle x = \int{\frac{1}{1 + 4y^2}\,dy}$

    $\displaystyle x = \frac{1}{2}\arctan{(2y)} + C$

    $\displaystyle x - C = \frac{1}{2}\arctan{(2y)}$

    $\displaystyle 2(x - C) = \arctan{(2y)}$

    $\displaystyle 2y = \tan{2(x - C)}$

    $\displaystyle y = \frac{1}{2}\tan{2(x - C)}$.


    You are told that $\displaystyle y(0) = 0$, so

    $\displaystyle 0 = \frac{1}{2}\tan{2(0 - C)}$

    $\displaystyle 0 = \frac{1}{2}\tan{(-C)}$

    $\displaystyle \tan{(-C)} = 0$

    $\displaystyle C = 0$.



    Thus $\displaystyle y = \frac{1}{2}\tan{(2x)}$.
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