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Simple homogenous 2nd order Diff EQ, but i'm stuck on the process!

  1. Feb 8, 2006 #1
    Hello everyone, we just started 2nd order differentials, and i was loooking at his example and it made senes but now i'm doing the hw and i'm stuck.

    Here is the problem:
    Find y as a function of t if
    y'' - 3y' = 0,
    y(0) = 9, y(1) = 7 .
    y(t) =?

    Well there is my work
    [​IMG]
     
  2. jcsd
  3. Feb 8, 2006 #2

    benorin

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    you have it right, just need to solve for c1 and c2 from this system (same as you have):

    [tex]c_1+c_2=9[/tex]
    [tex]c_1+c_2e^3=7[/tex]
     
  4. Feb 8, 2006 #3

    benorin

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    Use matrix methods: the system is

    [tex]\left(\begin{array}{cc}1&1\\1&e^3\end{array}\right) \left(\begin{array}{cc}c_1\\c_2\end{array}\right) =\left(\begin{array}{cc}9\\7\end{array}\right)[/tex]
     
  5. Feb 8, 2006 #4

    benorin

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    multipling on the left by the inverse of the coefficient matrix gives

    [tex]\left(\begin{array}{cc}c_1\\c_2\end{array}\right) = \frac{1}{(1)(e^3)-(1)(1)} \left(\begin{array}{cc}e^3&-1\\-1&1\end{array}\right) \left(\begin{array}{cc}9\\7\end{array}\right) = \frac{1}{e^3-1} \left(\begin{array}{cc}9e^3-7\\-9+7\end{array}\right) = \left(\begin{array}{cc}\frac{9e^3-7}{e^3-1}\\-\frac{2}{e^3-1}\end{array}\right)[/tex]

    so [tex]c_1=\frac{9e^3-7}{e^3-1}\mbox{ and }c_2=-\frac{2}{e^3-1}[/tex]
     
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