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anonymous

  • 4 years ago

y"+6y'+9=0 can someone help me to explain this ODE?

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  1. anonymous
    • 4 years ago
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    you can assume the solution is in the form of \[y=e^{\lambda x}\]then when you sub that in for y, you just need to solve for the lambda values.

  2. anonymous
    • 4 years ago
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    so you get labda = -3 twice right, so y = \[e^{-3x}\] + x*\[e^{-3x}\] right?

  3. anonymous
    • 4 years ago
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    Well, what you have in front of you is a very normal differential equation, I will assume you want the solution in R and not in C... So, here we go: The characteristic equation to this equation is \[r^2+6r+9=0 \] the discriminant is therefore \[\Delta=6^2-4*9=36-36=0\] So we have one solution \[r=\frac{-6}{2*9}=\frac{-1}{3}\] and the canonical solution for a differential equation in real values with a characteristic equation with only one solution is, finally \[y:t \rightarrow (A*t+B)*e^{\frac{-1}{3}*t} \] where A and B are constants determined by the initial conditions

  4. anonymous
    • 4 years ago
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    r should be -6/(2*1) no?

  5. anonymous
    • 4 years ago
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    not (2*9)?

  6. anonymous
    • 4 years ago
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    oh yes, small mistake... so you end up with r=-3 instead, my bad

  7. anonymous
    • 4 years ago
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    great thanks :)

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