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anonymous

  • 2 years ago

find the differential solution to the equation to which the given function is the general solution y=(A+Bx+Cx^2)e^(2x)

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  1. Loser66
    • 2 years ago
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    I don't understand the question. What are we supposed to do?

  2. Loser66
    • 2 years ago
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    I guess, you want to find out the function? if so, from your general solution, I expand it to \(y= Ae^{2x}+ Bxe^{2x}+Cx^2e^{2x}\) that shows you have \(\lambda =2\) triple roots so, the characteristic equation is \((\lambda -2)^3 =0\) and then \(= \lambda^3 -6\lambda^2+12\lambda -8 =0\) that gives us the original function is y''' -6y''+12y' -8y =0 that's all I know. Hope this help

  3. anonymous
    • 2 years ago
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    i figure it out thank you i was suppose to find the equation it was y'''-6y''+8y'-8y=0

  4. Loser66
    • 2 years ago
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    12y' , not 8y'

  5. anonymous
    • 2 years ago
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    I just seen that thank you

  6. Loser66
    • 2 years ago
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    ok

  7. abb0t
    • 2 years ago
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    Wow. Congrats on reaching \(\sf \color{limegreen}{green}\) status. You're now part of the elite.

  8. Loser66
    • 2 years ago
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    hihihi.... just stay so long in this site and get "green" . no smart, no knowledge

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