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Thread: differential equation

  1. #1
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    differential equation

    a)for what values of r does the function $\displaystyle y=e^{rx}$ satisfy the differential equation $\displaystyle 2y''+y'-y=0$

    b)If $\displaystyle r_1$ and $\displaystyle r_2$ are the values of r the were found in part a, show that every member of the family of functions is $\displaystyle y=a e^{r_1x} +b e^{r_2 x}$ is also a solution.

    so I took the first and second derivative getting $\displaystyle y'=re^{rx}$ and $\displaystyle y''=r^2 e^{rx}$.

    Therefore:

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

    $\displaystyle \Rightarrow e^{rx} (2r^2+r+1) = 0$

    I can't factor that any farther so I don't know how to find the roots.
    Last edited by mr fantastic; Jul 12th 2009 at 04:53 PM. Reason: Fixed the latex.
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  2. #2
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    a)for what values of r does the function $\displaystyle y=e^(rx)$ satisfy the differential equation $\displaystyle 2y''+y'-y=0$
    Try solving $\displaystyle 2m^{2}+m-1=0$
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  3. #3
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    Quote Originally Posted by superdude View Post
    a)for what values of r does the function $\displaystyle y=e^{rx}$ satisfy the differential equation $\displaystyle 2y''+y'-y=0$

    b)If $\displaystyle r_1$ and $\displaystyle r_2$ are the values of r the were found in part a, show that every member of the family of functions is $\displaystyle y=a e^{r_1x} +b e^{r_2 x}$ is also a solution.

    so I took the first and second derivative getting $\displaystyle y'=re^{rx}$ and $\displaystyle y''=r^2 e^{rx}$.

    Therefore:

    $\displaystyle 2r^2e^{rx} + re^{rx} {\color{red}+} e^{rx} = 0$

    $\displaystyle \Rightarrow e^{rx} (2r^2+r {\color{red}+} 1) = 0$

    I can't factor that any farther so I don't know how to find the roots.
    Very careless. The red plus is clearly wrong.
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  4. #4
    Super Member Random Variable's Avatar
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    Since $\displaystyle e^{rx}$ can never be zero, $\displaystyle 2r^{2}+r-1$ must be zero.
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