Question

Just need this one checked. Solve \[y'''-2y'+y=te^t\]
Char eq. \[r^3-2r+1=0\]\[(r-1)(r^2+r-1)=0\]\[r=1,~-.5(1+\sqrt 5),~ .5(\sqrt 5-1)\]

Answer 1

then my homogeneous solution is\[ y= c_1e^t+c_2e^{-.5t(1+\sqrt 5)}+c_3e^{.5t(\sqrt5-1)}\]

Answer 2

Do you agree to this point?

Answer 3

use undetermined coeff of variation of paramets for the particular solution

Answer 4

or* variation of parameters

Answer 5

but do you agree to that point?

Answer 6

I din't muff anything up through there?

Answer 7

ok, so for MOUC \(y_p(t)=(t-1)e^t\)

Answer 8

yea?

Answer 9

@amistre64 do you mind checking this one so far?

Answer 10

Apparently I am wrong

Answer 11

the partiticular sln should be of the form (At^2+Bt)e^t

Answer 12

\[y'''-2y'+y=te^t\]
\[(e^{rx})'''-2(e^{rx})'+(e^{rx})=0\]
\[r^3(e^{rx})-2r(e^{rx})+(e^{rx})=0\]
\[r^3-2r+1=0\]
\[r=1,\frac12(-1\pm\sqrt{5}) \]
\[y_h=c_0exp(x)+c_1exp(\frac{-1-\sqrt{5}}{2}x)+c_2exp(\frac{-1+\sqrt{5}}{2}x)\]

Answer 13

http://www.wolframalpha.com/input/?i=y%27%27%27-2y%27%2By%3Dte%5Et

Answer 14

x .. t... meh

Answer 15

it's ok,

Answer 16

I got what you meant

Answer 17

ever work a wronskian?

Answer 18

yea, but I don't have two solutions

Answer 19

ohhhhhhhH!H!!H!HH!H!!HH!H!!

Answer 20

you have a degree 3 and 3 soluitions

Answer 21

ce^t and ce^{.5(sqrt5-1)t

Answer 22

that just skipped over my head. That will be faster. Variation of parameters it is

Answer 23

thanks!

Answer 24

itll be a 3x4 matrix i believe. good luck :)

Answer 25

nah, we only need two of em

Answer 26

right?

Answer 27

my memory says no.

if we has a 2nd order y'' then 2 yes

we have a y''' order so its a 3x3 setup

Answer 28

we have 3 linearly independant terms in the homogenous

Answer 29

ew. Then I feel MOUC would be better in this instance

Answer 30

you dont know the 3x3 determinant shortcut? pretty much reduces to the 2x2 shortcut in a way; multiplicaiton of diags

Answer 31

I know, but it's not pretty and a mess to integrate

Answer 32

ok, ive just never come across MOUS before so i wouldnt know how that operates

Answer 33

method of undertermined coefficients

Answer 34

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

Answer 35

i call that variation of parameters ... of course my vocab in this area migh tbe rusty

Answer 36

variation of parameters uses integrals and wronskian

Answer 37

the answer to this is ridiculous

Answer 38

\[y_p=Af+Bg+Ch\]
\[y'_p=Af'+Bg'+Ch'+(A'f+B'g+C'h:=0)\]
\[y''_p=Af''+Bg''+Ch''+(A'f'+B'g'+C'h':=0)\]
\[y'''_p=Af'''+Bg'''+Ch'''+(A'f''+B'g''+C'h'':=te^t)\]

waaaaiiiittt a minute, we have constant coefficients so isnt this a bit much for it?

even tho this should still pan out ...

Answer 39

http://www.wolframalpha.com/input/?i=y%E2%80%B2%E2%80%B2%E2%80%B2%E2%88%922y%E2%80%B2%2By%3Dt+e%5Et

looks fine to me

Answer 40

http://www.wolframalpha.com/input/?i=y%E2%80%B2%E2%80%B2%E2%80%B2%E2%88%922y%E2%80%B2%2By%3Dt+e%5Et+%2C+y%280%29%3D0+%2C+y%27%280%29%3D0%2C+y%27%27%280%29%3D0

Answer 41

With initial conditions

Answer 42

i just pooed myself :)

Laplace transform maybe? since we have all those ynots

Answer 43

that may work better

Answer 44

that will work better :D

Answer 45

you didnt include the IVP stuff to start with or i might have suggested that ... well about the same time but still...

Answer 46

well, I shall do it now

Answer 47

so \[L(y)=\frac{1}{(s-1)^2(s^3-2s+1)}\]

Answer 48

yea?

Answer 49

so then \[L(t)=\frac{1}{(s-1)^3} *\frac{1}{s^2+s-1}

\]

Answer 50

\[s^3 L\{y\}+L\{y\}-2sL\{y\}=\frac{1}{(s-1)^2}\]

Answer 51

you seem to be doing fine :)

Answer 52

My issue is how to do the inverse L(t)

Answer 53

complete the squuare on the s^2+s+1 .... and i think your going to decomp into a sum of fraction right?

Answer 54

uhm, I know the roots of that, but the issue is how do I do the inverse L(t) for 3 things multiplied together

Answer 55

like this is the final answer http://www.wolframalpha.com/input/?i=inverse+Laplace+transform+%281%2F%5B%28s-1%29%5E3%28s%5E2%2Bs-1%29%5D%29

Answer 56

thats definiantly a better siplification than the other output

Answer 57

yea, but how do I get it....

Answer 58

\[\frac{1}{(s-1)^3(s^2+s+1)}=\frac{As^2+Bs+C}{(s-1)^3}+\frac{Ms+N}{(s^2+s+1)}\]

Answer 59

ah

Answer 60

thank you

Answer 61

yw

Answer 62

I got it from here now