Question

need help starting out: use the method of variation of parameters to determine the solution to the differential equation: y''' -3y'' + 4y = e^(2x)

Answer 1

first thing is to find the complimentary solution

Answer 2

so far i have....

Answer 3

you can just use undetermined coefficients...right?

Answer 4

it says variation of parameters, so...

Answer 5

oh wait...it said to use variation specifically

Answer 6

(D^3 -3D^2 +4)y = e^2x. that factored the (d+1)(d-2)^2 = e^2x

Answer 7

ohhh so this is what you were asking...

Answer 8

you have to change D into m first then factor like algebra

Answer 9

yea

Answer 10

and also make the right side = 0

Answer 11

you first want the homogeneous solution, so all that business should be =0

Answer 12

beat me too it

Answer 13

\m/

Answer 14

yes, then what do I do. what would (d-2)^2 become? e^(4x^2)?

Answer 15

what are you doing? im getting confused..

Answer 16

i am trying to find the fundamental solutions right now

Answer 17

set*

Answer 18

ohh you're solving for roots huh

Answer 19

(d-2)^2 means root of 2 with multiplicity 2 right?

Answer 20

or at least the general ssolution

Answer 21

when you have a root that's like \[m = a^2\]
the solution would be \[\large C_1 e^{ax} + C_2x e^{ax}\]
does that help?

Answer 22

so i would have 3 fundamental solutions: { e^-x , e^2x , and xe^2x}??

Answer 23

dont forget the constants

Answer 24

\[\huge y_c = c_1 e^{-x} + c_2 e^{2x} + c_3 x e^{2x}\]

Answer 25

oh, well then I would have ......YUP! that's what i was about to type... but maybe not as pretty as that

Answer 26

haha :p now do yp

Answer 27

can i not use Kramer's rule now?

Answer 28

derive twice then use matrices?

Answer 29

[g(x)W[y2,y3](x)] / [W[y1,y2,y3](x)]

Answer 30

they do that in variation of parameters? o.O all i know is you copy the yc and derive thrice (since third order differential equation) and then use matrices (but i just use systems of equations)

Answer 31

let me try to derive 3x's then.... oye...

Answer 32

but seriously...idk this "kramer's rule" you're saying..

Answer 33

it's okay

Answer 34

I guess you would use cramer's rule here, though I'm really only used to doing this for second-order problems

Answer 35

that's why i was saying systems would be nicer :C

Answer 36

.....i hate this pellet. it's waaay too confusing

Answer 37

HA! pellet was automatically filled in for sh*t

Answer 38

it takes time..especially if you're oriented with one method only

Answer 39

well, my teacher tries to give us at least one or two methods. but he ends up rushing himself and getting the problem wrong himslef...