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Thread: Seperable Differential Equation

  1. #1
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    Seperable Differential Equation

    Solve , using separation of variables, given the inital condition .

    I am not sure how to do this one. I tried to use trig substitution but it didnt work so I am not sure if that was the right path or not.
    Thanks
    AC
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  2. #2
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    Quote Originally Posted by Casas4 View Post
    Solve , using separation of variables, given the inital condition .

    I am not sure how to do this one. I tried to use trig substitution but it didnt work so I am not sure if that was the right path or not.
    Thanks
    AC
    $\displaystyle (t^2 + 16)\,\frac{dx}{dt} = x^2 + 16$

    $\displaystyle \left(\frac{1}{x^2 + 16}\right)\frac{dx}{dt} = \frac{1}{t^2 + 16}$

    $\displaystyle \int{\left(\frac{1}{x^2 + 16}\right)\frac{dx}{dt}\,dt} = \int{\frac{1}{t^2 + 16}\,dt}$

    $\displaystyle \int{\frac{1}{x^2 + 16}\,dx} = \int{\frac{1}{t^2 + 16}\,dt}$

    $\displaystyle \frac{1}{4}\arctan{\frac{x}{4}} + C_1 = \frac{1}{4}\arctan{\frac{t}{4}} + C_2$

    $\displaystyle \frac{1}{4}\arctan{\frac{x}{4}} = \frac{1}{4}\arctan{\frac{t}{4}} + C_2 - C_1$

    $\displaystyle \arctan{\frac{x}{4}} = \arctan{\frac{t}{4}} + C$, where $\displaystyle C = 4(C_2 - C_1)$

    $\displaystyle \frac{x}{4} = \tan{\left(\arctan{\frac{t}{4}} + C\right)}$

    $\displaystyle x = 4\tan{\left(\arctan{\frac{t}{4}} + C\right)}$


    And using the initial condition $\displaystyle x(0) = 4$

    $\displaystyle 4 = 4\tan{\left(\arctan{\frac{0}{4}} + C\right)}$

    $\displaystyle 4 = 4\tan{C}$

    $\displaystyle 1 = \tan{C}$

    $\displaystyle C = \arctan{1} = \frac{\sqrt{2}}{2}$.


    Therefore $\displaystyle x = 4\tan{\left[\arctan{\left(\frac{t}{4}\right)} + \frac{\sqrt{2}}{2}\right]}$
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  3. #3
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    Yes , then use the trig. formula

    $\displaystyle \tan(A + B ) = \frac{ \tan(A) + \tan(B) }{ 1- \tan(A) \tan(B) }$


    $\displaystyle x = 4 \frac{ 1 + t/4}{1- t/4} = 4 \frac{ 4 + t}{4-t} $
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