Question

D^4=a^4,where D=d/dx
plzzzzz....find the four roots for D.(proceed this way D^4=(1)a^4
D=(1)^1/4 *a
then use cis form in the place of 1.

Answer 1

is a constant??

Answer 2

yup

Answer 3

D^3 = a^4t + k_1
D^2 = a^4t^2/2 + k_1t + k_2
...
...
integrate with t ...

Answer 4

D^4=(cis 0)^1/4 *a

Answer 5

do you have answer??

Answer 6

yes......
a,-a,ia,-ia

Answer 7

oh,
\[ D^4 - a^4 = 0\]
\[ (D^2 - a^2)(D^2 + a^2) = 0\]

Answer 8

do u remember any formula like this....
cis(2k pi +1)theta /1.......something like this?

Answer 9

Something like this??
\[ (D^2 + a^2)y = 0 => y = c_1 \cos(ax) + c_2 \sin(ax)\]

Answer 10

if D^6=-1*a^6 ,then?

Answer 11

is
\[ D^2 = \frac{d^2y}{dx^2} \text { or } D^2 = (dy/dx)^2\]

Answer 12

hmn.

Answer 13

D^2=d^2/dx^2

Answer 14

first one right??

Answer 15

yes

Answer 16

oh na!!