amorfide
  • amorfide
When getting the complimentary function, does it matter how I write it? Obviously you get two distinct roots, m=3 and m=2 but does it matter which is raised to what power?
Mathematics
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jamiebookeater
  • jamiebookeater
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amorfide
  • amorfide
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amorfide
  • amorfide
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amorfide
  • amorfide
does it matter what way? or are both right?

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anonymous
  • anonymous
The complementary homogeneous equation is $$y''-5y'+6y=0$$To determine the complementary solution, presume \(y=e^{\lambda t}\), and thus \(y'=\lambda e^{\lambda t},y''=\lambda^2e^{\lambda t}\). Substitute:$$\lambda^2e^{\lambda t}-5\lambda e^{\lambda t}+6e^{\lambda t}=0\\e^{\lambda t}\left(\lambda^2-5\lambda+6\right)=0\quad\implies\lambda^2-5\lambda+6=0\\\lambda^2-5\lambda+6=(\lambda-3)(\lambda-2)=0\quad\implies\lambda=2,3$$... and thus we know our equation can be a linear combination of \(e^{2t},e^{3t}\), i.e. $$y=Ae^{2t}+Be^{3t}$$Does it matter which term I put first? Which coefficient I label A and which I label B? No, so long as I'm consistent.
amorfide
  • amorfide
so Ae^3x + Be^2x is okay instead of Ae^2x + Be^3x ?
amorfide
  • amorfide
@oldrin.bataku
anonymous
  • anonymous
\(y=Ae^{2t}+Be^{3t}\) and \(y=Be^{3t}+Ae^{2t}\) are equivalent, aren't they? and it doesn't matter which coefficient you label \(A,B\) as long as you remain consistent.
amorfide
  • amorfide
I would like a yes or no answer, please?
anonymous
  • anonymous
I have very clearly said it was fine twice now.
amorfide
  • amorfide
You never, but now you have. Thank you!

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