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

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