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30 March, 19:24

The indicated function y1 (x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section 4.2, y2 = y1 (x) e-∫P (x) dx y 2 1 (x) dx (5) as instructed, to find a second solution y2 (x). y'' - 8y' + 16y = 0; y1 = e4x

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  1. 30 March, 23:00
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    y2 = C1xe^ (4x)

    Step-by-step explanation:

    Given that y1 = e^ (4x) is a solution to the differential equation

    y'' - 8y' + 16y = 0

    We want to find the second solution y2 of the equation using the method of reduction of order.

    Let

    y2 = uy1

    Because y2 is a solution to the differential equation, it satisfies

    y2'' - 8y2' + 16y2 = 0

    y2 = ue^ (4x)

    y2' = u'e^ (4x) + 4ue^ (4x)

    y2'' = u''e^ (4x) + 4u'e^ (4x) + 4u'e^ (4x) + 16ue^ (4x)

    = u''e^ (4x) + 8u'e^ (4x) + 16ue^ (4x)

    Using these,

    y2'' - 8y2' + 16y2 =

    [u''e^ (4x) + 8u'e^ (4x) + 16ue^ (4x) ] - 8[u'e^ (4x) + 4ue^ (4x) ] + 16ue^ (4x) = 0

    u''e^ (4x) = 0

    Let w = u', then w' = u''

    w'e^ (4x) = 0

    w' = 0

    Integrating this, we have

    w = C1

    But w = u'

    u' = C1

    Integrating again, we have

    u = C1x

    But y2 = ue^ (4x)

    y2 = C1xe^ (4x)

    And this is the second solution
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