Question

Can someone please help me!
Find a particular solution of the differential equation
(-5/4)y''−2y'+y=2xe^(4x)
using the Method of Undetermined Coefficients (primes indicate derivatives with respect to x).

Answer 1

constant coefficients eh

Answer 2

yea i ended up getting (20/729)e^(4x)-(2/27)xe^(4x) and its wrong!

Answer 3

I tried it 3 times and got the same answer

Answer 4

-5/4 r^2 -2r +1 = 0 is the characteristic eq which makes this thing look like it might go ugly

Answer 5

but we only need to solve for the particular solution

Answer 6

\[r=\frac{2\pm\sqrt{4-4(-5/4)}}{-5/2}\]
\[r=\frac{2\pm\sqrt{4+5}}{-5/2}\]
\[r=\frac{2\pm3}{-5/2}\to\ \frac{4\pm 6}{-5}\]

Answer 7

cant solve for p without c can you?

Answer 8

I thought you didntreally need to solve for the complimentary solution unless you were solving for the whole solution

Answer 9

but i may be wrong

Answer 10

i cant verify that, i always remember doing it this way :)

Answer 11

ok

Answer 12

\[\large y_c=c_1exp({-\frac45 x})+c_2exp({\frac65 x})\]

Answer 13

what I did was that i found yp=Ae^4x +Bxe^4x then i derived that and derived it again then I plugged it into the original equation and then solved for the coefficients A and B

Answer 14

your way way might be right, its been awhile since i tried to figure out the specifics

Answer 15

id have to read up on undetermines to be sure tho

Answer 16

but my answer ended up being wrong so please continue on ith hat you were doing

Answer 17

well, my way might get to an answer, but its prolly the wrong method its asking for :)

Answer 18

true but if I know the correct answer then I might be able to figure out where I went wrong, so if you can, please continnue

Answer 19

\[\large y_p=Aexp({-\frac45 x})+Bexp({\frac65 x})\]
\[\large y'_p=\cancel{A'exp({-\frac45 x})+B'exp({\frac65 x})}^0\]\[\ \ \ \ \large -A\frac45 exp({-\frac45 x})+B\frac65exp({\frac65 x})\]

\[\large y''_p=-A'\frac45 exp({-\frac45 x})+B'\frac65exp({\frac65 x})\]\[\large \ \ \ \ +A\frac{16}{25} exp({-\frac45 x})+B'\frac{36}{25}exp({\frac65 x})\]

then its alot of fractioning math lol

Answer 20

im outof time for today, but good luck.

to double chk your answers, use the wolf.

http://wolframalpha.com

Answer 21

http://www.wolframalpha.com/input/?i=%28-5%2F4%29y%27%27%E2%88%922y%27%2By%3D2xe%5E%284x%29

looks like you got a -2x/27 e^4x

Answer 22

thanks