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Second-Order Homogeneous Linear Equation

  1. Apr 2, 2005 #1
    I have this intial value problem: y''-4y'-5y=0, y(1)=0, y'(1)=2. My AUX equation is r^2-4r-5=0. I factor and get r=5, r=-1 and my equation becomes y(x)=C1e^(5x)+C2e^(-1x) (C1 and C2 are constants). I took the derivative of y(x) and then tried to use my initial value's to solve for C1 and C2. I got C1e^5+C2e^(-1)=0 and 5C1e^5-C2e^(-1)=2. For some reason when I solve for C1 or C2 and put them back in the equations it doesn't check out. I'd appreciate any advice to help me find what C1 and C2 are. Thanks!!!
  2. jcsd
  3. Apr 2, 2005 #2
    [tex] C_1e^5 + C_2e^{-1} = 0, \ 5C_1e^5 - C_2e^{-1} = 2 \Longrightarrow 6C_1e^5 = 2 \Longrightarrow C_1 = \frac{1}{3}e^{-5}[/tex]


    [tex] C_1e^5 + C_2e^{-1} = \frac{1}{3}+C_2e^{-1}=0 \Longrightarrow C_2 = -\frac{e}{3}.[/tex]
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