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mash

  • 3 years ago

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

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  1. mash
    • 3 years ago
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    A and B are 2 arbritary constants

  2. John_ES
    • 3 years ago
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    I 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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