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Calculus

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Show that both functions y1=e^(-2x) and y2=xe^(-2x) are solutions to the differential equation y^''+4y^'+4y=0

  • Calculus -

    what's the troub;e? Just plug and chug

    y1 = e^(-2x)
    y1' = -2e^(-2x)
    y1" = 4e^(-2x)

    y"+4y'+4 = 0

    y2 = xe^(-2x)
    y2' = (1-2x) e^(-2x)
    y2" = (4x-4) e^(-2x)

    y"+4y'+4y = 0

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