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Math Help - Differential Equation help...

  1. #1
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    Differential Equation help...

    Hello, I'm a newbie when it comes to differential equations and my professor assigned some problems that we never went over in class. Hopefully someone can help...

    y' = 2y; y(x) = Ce^(2x), y(0) = 3.

    Here are the directions: First verify that y(x) satisfies the given differential equation. Then determine a value of the constant C so that y(x) satisfies the given initial condition.

    I have to do about 10 of these, so if anyone could go through the steps on how to solve this, I will greatly appreciate it. Thanks for your time...
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  2. #2
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    \frac{dy}{dx} = 2y

    \frac{1}{y}\,\frac{dy}{dx} = 2

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

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

    \ln{|y|} + C_2 = 2x + C_1

    \ln{|y|} = 2x + C_1 - C_2

    |y| = e^{2x + C_1 - C_2}

    |y| = e^{C_1 - C_2}e^{2x}

    y = \pm e^{C_1 - C_2}e^{2x}

    y = Ce^{2x} where C = \pm e^{C_1 - C_2}.


    Now use the inital condition y(0) = 3

    3 = Ce^{2\cdot 0}

    3 = C.


    Therefore y = 3e^{2x}.
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