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gerryliyana Group Title

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

  • 2 years ago
  • 2 years ago

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  1. gerryliyana Group Title
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    @blues

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

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

    • 2 years ago
  4. gerryliyana Group Title
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    thank you :)

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

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

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

    • 2 years ago
  8. UnkleRhaukus Group Title
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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

    • 2 years ago
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