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Math Help - 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

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

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

     <br />
sec^2(2x+ C) = 1 + 4(\frac{1}{2}tan(2x+C))^2<br />

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

    sec^2(2x+ C) = 1 + tan^2(2x+C))

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

    y(0) = 0 so

    sec^2(2x+C) = 1 + tan^2(2x+C))

    sec^2(2(0) + C) = 1 + tan^2(2(0) + C)

    sec^2(C) = 1 + tan^2(C)

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

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

    cos^2(C) = 1 + sin^2(C)

    C = n* \pi

    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

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

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

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

    y(0)=0 is used to solve for C.
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  3. #3
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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

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

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

     <br />
sec^2(2x+ C) = 1 + 4(\frac{1}{2}tan(2x+C))^2<br />

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

    sec^2(2x+ C) = 1 + tan^2(2x+C))

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

    y(0) = 0 so

    sec^2(2x+C) = 1 + tan^2(2x+C))

    sec^2(2(0) + C) = 1 + tan^2(2(0) + C)

    sec^2(C) = 1 + tan^2(C)

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

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

    cos^2(C) = 1 + sin^2(C)

    C = n* \pi

    cos^2(0) = 1 + sin^2(0)

    1 = 1

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

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

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

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

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

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

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

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


    You are told that y(0) = 0, so

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

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

    \tan{(-C)} = 0

    C = 0.



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