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Thread: Using seperation of variables?

  1. #1
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    Using seperation of variables?

    What is the procedure or what are the steps to be carried out here:
    $\displaystyle \frac{X\prime}{X}=\frac{1}{4}(\frac{Y}{Y\prime})=\ lambda$
    The answer is given as $\displaystyle X=e^{\lambda\alpha}$ and $\displaystyle Y=e^{\frac{\beta}{4\lambda}}$
    I don't know how to get to the answer, i have done seperation of variables, but not when i have a derivative as a denominator!
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by punkstart View Post
    What is the procedure or what are the steps to be carried out here:
    $\displaystyle \frac{X\prime}{X}=\frac{1}{4}(\frac{Y}{Y\prime})=\ lambda$
    The answer is given as $\displaystyle X=e^{\lambda\alpha}$ and $\displaystyle Y=e^{\frac{\beta}{4\lambda}}$
    I don't know how to get to the answer, i have done seperation of variables, but not when i have a derivative as a denominator!
    This produces 2 equations

    $\displaystyle X^{ \prime } - \lambda X = 0$

    and

    $\displaystyle \lambda Y^{ \prime } - \frac{Y}{4} = 0$

    This can become

    $\displaystyle Y^ { \prime } - \frac{ Y }{ \lambda 4} = 0$

    To solve both,

    let $\displaystyle X=e^{rx}$

    let $\displaystyle Y=e^{qy} $

    Therefore the first equation becomes

    $\displaystyle re^{rx} - \lambda e^{rx} = 0$

    $\displaystyle r= \lambda $

    Hence,

    $\displaystyle X=C_1 e^{ \lambda x } $

    Of course the co-efficient is just a constant to represent the general solution. Which can be re-written in the form that they gave you.

    The same follows for the Y equation!
    Last edited by AllanCuz; Apr 19th 2010 at 03:17 PM.
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