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Thread: Solve d^2y/dx^2 + 4y = 0?

  1. #1
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    Thumbs up Solve d^2y/dx^2 + 4y = 0?

    Hi there,

    $\displaystyle (d^2y)/(dx^2) + 4y = 0$ is obviously the second derivative.
    What I know is that if its $\displaystyle dy/dx$ I must separate the dy and only the y to one side , and the x's and any number attached to the y along with dx to the other side ( other side of the = sign ). After that is done, we integrate both sides to find the answer. Okay sweet, but how do we apply that on the second derivative ?

    Thank you
    Last edited by ZaZu; Oct 1st 2009 at 11:45 AM.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by ZaZu View Post
    Hi there,

    $\displaystyle (d^2y)/(dx^2) + 4y = 0$ is obviously the second derivative.
    What I know is that if its $\displaystyle dy/dx$ I must separate the dy and only the y to one side , and the x's and any number attached to the y along with dx to the other side ( other side of the = sign ). After that is done, we integrate both sides to find the answer. Okay sweet, but how do we apply that on the second derivative ?

    Thank you
    you can see this http://www.mathhelpforum.com/math-help/differential-equations/38182-differential-equations-tutorial.html


    in the second order equation let $\displaystyle y=e^{rx}$ then find r

    $\displaystyle r^2e^{rx} +4e^{rx} = 0 $

    $\displaystyle e^{rx}(r^2 + 4)=0\Rightarrow r=\mp 2i $ two complex root

    see the link for more information ...
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  3. #3
    MHF Contributor Matt Westwood's Avatar
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    Quote Originally Posted by ZaZu View Post
    Hi there,

    $\displaystyle (d^2y)/(dx^2) + 4y = 0$ is obviously the second derivative.
    What I know is that if its $\displaystyle dy/dx$ I must separate the dy and only the y to one side , and the x's and any number attached to the y along with dx to the other side ( other side of the = sign ). After that is done, we integrate both sides to find the answer. Okay sweet, but how do we apply that on the second derivative ?

    Thank you
    One way of doing it is by putting $\displaystyle p = \frac {dy}{dx}$.

    This leads to $\displaystyle \frac {d^2y}{dx^2} = \frac {dp}{dx} = \frac {dp}{dy} \frac {dy}{dx}$ (by the chain rule).

    So $\displaystyle \frac {d^2y}{dx^2} = p \frac {dp}{dy}$

    and so we can rewrite it:
    $\displaystyle p \frac {dp}{dy} + 4 y = 0$
    which is clearly separable.

    So you can solve for $\displaystyle p$, and this will then be a 1st order equation which (when you've done the algebra) you will find also separates.
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  4. #4
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    Thank you very much guys

    Greatly appreciated
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