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I'm working on solving the following homogenous equation:

$$ y'' - 8y' + 16y = 0 $$

Seems like a straight forward $y=e^{rx}$ substitution and then solve for r1 and r2:

$$ y=e^{rx}=0 $$ $$ y=re^{rx}=0 $$ $$ y=r^2e^{rx}=0 $$

$$ r^2e^{rx} - 8re^{rx} + 16e^{rx}=0 $$ $$ e^{rx}(r^2-8r+16)=0 $$

Since $e^{rx}$ can't equal zero:

$$ r^2-8r+16 = 0 $$ $$ (r-4)(r-4)=0 $$ $$ r_{1,2} = 4 $$

How do I express the generaal solution from here? I've tried the following but it is incorrect:

$$ y=c_1e^{4x}+c_2e^{4x} $$

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Because of the repeated root, we have to express it as:

$$ y(x) =c_1e^{4x}+c_2 x e^{4x} $$

See Paul's Online Notes for theory as to why.

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  • $\begingroup$ Short and sweet :-) +1 $\endgroup$ – Namaste Oct 7 '13 at 0:41
  • $\begingroup$ How are U feeling today? It was last week Monday that you saw the Doctor? $\endgroup$ – Namaste Oct 8 '13 at 14:55
  • $\begingroup$ @amWhy: Yes, last week was doc. I was still having effects throughout the weekend, so did light workout Sat and Sun. Feel about 95% today, so almost there. I know you are getting down to a few more days, so hope all is well with you too! Have a great day! Been trying to stay at bay from MSE also! $\endgroup$ – Amzoti Oct 8 '13 at 15:38

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