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 one year ago
how do you get from y=Ae^4x + Be^6x
to differential eqution f''(x) +10f(x) +24y=0
 one year ago
how do you get from y=Ae^4x + Be^6x to differential eqution f''(x) +10f(x) +24y=0

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mash
 one year ago
Best ResponseYou've already chosen the best response.0A and B are 2 arbritary constants

John_ES
 one year ago
Best ResponseYou've already chosen the best response.0I think you part from the differential equation. Then, you use the replacement, \[f(x)=C e^{rx}\] So, \[f''(x)+10f'(x)+24f(x)=0\Rightarrow r^2+10r+24=0\Rightarrow r_1=6, r_2=4\] Then, the general solution is, \[f(x)=Ae^{4x}+Be^{6x}\]
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