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gerryliyana

  • 2 years ago

Determine whether y(x) = (2e^-x) + (xe^-x) is a solution of y'' + 2y' + y = 0

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  1. gerryliyana
    • 2 years ago
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    @blues

  2. UnkleRhaukus
    • 2 years ago
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    \[y= 2e^{-x} + xe^{-x}\] \[y'=\] \[y''=\]

  3. UnkleRhaukus
    • 2 years ago
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    find the first and second derivatives , if \(y\) is a solution then \[y'' + 2y' + y = 0\]

  4. gerryliyana
    • 2 years ago
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    thank you :)

  5. UnkleRhaukus
    • 2 years ago
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    what do you get for the first and second derivative?

  6. gerryliyana
    • 2 years ago
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    yes of course.., y' (x) = -2e^(-x) + e^(-x) - xe^(-x) = -e^(-x) - xe^(-x) right??

  7. gerryliyana
    • 2 years ago
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    ok for y'' (x) = e^(-x) - e^(-x) + xe^(-x) = ex^-x

  8. UnkleRhaukus
    • 2 years ago
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    thats right but i think you typed the very last bit the wrong way around. now substitute these into \[y′′+2y′+y=0\] \[ (xe^{−x})+2(−e^{−x}−xe^{−x})+(2e−x+xe^{−x})=0\] if y is a solution then this will simplify to a statement that is always true

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