Question

if y1=cos4x is a solution to the DE: y''+16y=0. Find y2

Answer 1

Find a y2 that is also a solution

Answer 2

\[y_2=y_1\int{}{}\frac{e^\int{}{}{p(x)dx}}{y_1^2}\]

or use substitution y=u(x)y1

Answer 3

y=ucos4x

y'=-4usin4x+cos4xu'

Answer 4

it's supposed to reduce to a order 1 linear: y'+p(x)y=0 using the y=ux

Answer 5

y=uy1 i mean

Answer 6

the integral way you get a crazy integral

Answer 7

wait i might know what happened

Answer 8

ahh i wrote it down wrong on my paper but here it is right

Answer 9

y''=-16ucos4x-4u'sin4x+cos4xu''-4u'sin4x=-16ucos4x-8u'sin4x+cos4xu''

Answer 10

y''+16y=-16ucos4x-8u'sin4x+cos4xu''+16ucos4x=-8u'sin4x+cos4xu''

Answer 11

let w=u'

cos(4x)w'-8sin(4x)w=0

Answer 12

now it's in linear

Answer 13

\[w'\frac{-8sin(4x)}{cos(4x)}w=0\]

\[w'-8tan(4x)w=0\]

Answer 14

y2=sin(4x)

Answer 15

how'd you get it

Answer 16

\[e^{\int{}{}-8tan(4x)}\]

Answer 17

\[e^{\int{}{}-2tan(u)du}\]=
\[e^{-2(-ln|cos u|)}\]=
\[e^{ln|cosu^2|}\]=
\[cos^2u=cos^2(4x)\]

Answer 18

Just replace y by sin(4x) and y'' by -16 sin(4x)

Answer 19

that's not showing anyy work at all