Question

How would you solve this equation?

Answer 1

y\[y ^{4}+y \prime \prime \prime-3y \prime \prime-5y \prime-2y=0\]

Answer 2

ignore the first y on the 1st line.

Answer 3

\[D(D^2-3D-5)y+(y^4-2y)=0\]
solve the "D" to get the general solution and the the particular solution
lengthy one man

Answer 4

D = deifferential operator

Answer 5

unfortunately, I got like 3 or 4 more like these to go.

Answer 6

get the roots of the operator function gives you the exponential solutions

Answer 7

I got -1 and 2

Answer 8

there should be a "0" too

Answer 9

@amistre64 @satellite73

Answer 10

I assumed that they were repeated roots.

Answer 11

@Mertsj

Answer 12

hope one of those helps.

Answer 13

Thanks!

Answer 14

I hope that they answer, too.

Answer 15

yeah, electro did well, you solve the characteristic polynomial for the roots and then your solution has to do with c_n e^(r_n x), given the r_n is a root

Answer 16

I did and am gettin -1 and 2.

Answer 17

r^4 +r^3 - 3r^2 - 5r - 2 = 0

Answer 18

if you only have 2 roots out of the 4, you might want to modify the e^rx with an x :)

Answer 19

eg?

Answer 20

in other words, if you have a multiply root; k

then you modify it as such:\[c_1e^{kx}+c_2x~e^{kx}+c_3 x^2 e^{kx}+...+c_n x^{n-1}e^{kx}\]for n multiples of a root

Answer 21

are the roots multiples? or are there complex roots as well?

Answer 22

i believe that they are just multiples.

Answer 23

so ... 2,-1,-1,-1

Answer 24

well, would it be -1, -1, 2, and 2?

Answer 25

no, the -2 is a single root, there are 3 multiples of -1 as roots

Answer 26

pfft, the 2 is a single root :)

Answer 27

oh right, i forgot to use my brain. ok the?

Answer 28

then*

Answer 29

then just write up the solution using the modification of a multiple root

Answer 30

nevermind i think i got it. thanks! I really appreciacte it!

Answer 31

\[y=c_0e^{2x}+c_1e^{-x}+c_2xe^{-x}+c_3x^2e^{-x}\]
\[y=c_0e^{2x}+e^{-x}(c_1+c_2x+c_3x^2)\]

Answer 32

if your solution had used the x^m format, then modification are made by multiples of ln(x)

Answer 33

thanks!

Answer 34

can you help me with a few more?

Answer 35

good luck ;)

Answer 36

i can try

Answer 37

thanks, let me start a new post or should i give them out on here?

Answer 38

a new post would allow others to participate and provide their insights as well. freshens things up :)