Question

find the general solution of the given differential equation:

y''+4y'+4y=t^(-2)e^(-2t)

Answer 1

use the characteristic equation:
r^2+4r+4 = t^(-2)e^(-2t)

Answer 2

i meant
r^2 + 4r + 4 = 0

Answer 3

yes, and variation of parameters should work for the particular

Answer 4

which means you need the wronskian...

Answer 5

find the roots:
(r+2)^2 = 0
r = -2 <= repeated root, i don't remember what to do next lol

Answer 6

Turing continue lol :) Love ya man, u help me and others soo much!

Answer 7

i need the particular solution
yp

Answer 8

platonically lol, just wanted to say thanks :)

Answer 9

haha, likewise
repeated root mean you add an x to one of the solutions\[y_c=c_1e^{-2t}+c_2te^{-2t}\]do you know how to find the Wronskian?

Answer 10

or rather, do you know how to do variation of parameters at all?

Answer 11

thats what i got for yc

Answer 12

I'm having difficult finding the yp

Answer 13

for a differential equation \[ay''+by+cy=g(t)\]who's complimentary solution is given by\[y_c=c_1y_1+c_2y_2\]the formula for the particular is\[y_p=-y_1\int\frac{y_2g(t)}{W}dt+y_2\int\frac{y_1g(t)}Wdt\]where W is the wronskian

Answer 14

now do you know how to find the wronskian?

Answer 15

yes i got it thanks

Answer 16

so what part are you stuck on?

Answer 17

i couldn't find the wronskian

Answer 18

this is the wronskian\[\det\left[\begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right]=y_1y_2'-y_2y_1'\]that is a determinate above

Answer 19

we have\[y_1=e^{-2t}\]\[y_2=te^{-2t}\]so the wronskian is\[W=\det \left[\begin{matrix}e^{-2t} & te^{-2t} \\ -2e^{-2t} & e^{-2t}-2te^{-2t}\end{matrix}\right]=e^{-4t}\]good so far?

Answer 20

yup

Answer 21

thats helpful

Answer 22

so\[y_p=-y_1\int\frac{y_2g(t)}{W}dt+y_2\int\frac{y_1g(t)}Wdt\]\[=-e^{-2t}\int\frac{te^{-2t}\cdot t^{-2}e^{-2t}}{e^{-4t}}dt+te^{-2t}\int\frac{e^{-2t}\cdot t^{-2}e^{-2t}}{e^{-4t}}dt\]

Answer 23

\[y_p=-e^{-2t}\int\frac{dt}t+te^{-2t}\int\frac{dt}{t^2}=-e^{-2t}\ln t-e^{-2t}\]\[y_p=-e^{-2t}(\ln t+1)\]questions?

Answer 24

i got the same thing but not 1 thou

Answer 25

I factored e^(-2t) out at the end is all

Answer 26

-e^-2t(lnt)

Answer 27

that's not right, what happened to your other integral?

Answer 28

your right