Question

Find general solution to differential equation.
16y'' +8y' +y = 0

Answer 1

Characteristic equation is 16r^2 + 8r +1 = 0

Answer 2

I thought to use quadratic to solve...

Answer 3

sorry, actually it is factorable.

Answer 4

yep

Answer 5

SO is general solution then y = C1e^-t/4 + C2e^-t/4 ?

Answer 6

nope, you need two independent solutions

Answer 7

you need to find another solution which doesnt look like e^-t/4

Answer 8

ok.....any tips where to start.

Answer 9

say the second independent solution is \(u(t) e^{\frac{-t}{4}}\)

Answer 10

\(y = u* e^{\frac{-t}{4}}\)
\(y' = ?\)
\(y'' = ?\)

Answer 11

just a sec..

Answer 12

y' = v'e^-t/4 - 1/4 *e^-t/4 * v

y'' = v''e^-t/4 - 1/4 *e^/t/4 *v' + 1/8 *e^-1/4 * v -1/4 *e^-t/4 v'

Answer 13

yes combine v, v' and v'' terms and write it in a readable form lol

Answer 14

you mean combine the two v' terms I am assuming.

Answer 15

Yep

\(y = u* e^{\frac{-t}{4}} \)

\(y' = u' e^{\frac{-t}{4} } - u\frac{1}{4} e^{\frac{-t}{4}} \)

\(y'' = u'' e^{\frac{-t}{4} } - u'\frac{1}{4}e^{\frac{-1}{4}} - u' \frac{1}{4} e^{\frac{-t}{4}} + u \frac{1}{16} e^{\frac{-t}{4}} \)

\(~~~~ = u'' e^{\frac{-t}{4} } - u'\frac{1}{2}e^{\frac{-1}{4}} + u \frac{1}{16} e^{\frac{-t}{4}} \)

Answer 16

see if that looks okay

Answer 17

same thing I wrote...thanks I got it from here.

Answer 18

http://www.wolframalpha.com/input/?i=16%28u%28t%29e%5E%28-t%2F4%29%29%27%27+%2B+8%28u%28t%29e%5E%28-t%2F4%29%29%27+%2B+%28u%28t%29e%5E%28-t%2F4%29%29

Answer 19

you should get \(u(t) = t\)

Answer 20

let me post a photo of what I have....

Answer 21

okie

Answer 22

Not sure what to do at the end ); So is it because it is one repeated soltuion to original characteristic that we have to find a unique seperate solution this way?

Answer 23

If I remember right, if you have something like (m+1)^2 the general solution is

\[\huge y = C_1 e^{-m}+ C_2Xe^{-m}\]
or something like that. What I mean is if you have a square, you just put something like an X with a higher degree. I may be remembering wrong though

Answer 24

Sorry photo almost too small to read.

Answer 25

Thats what I thought for a minute also but not sure....

Answer 26

adding up should give u :

\(16 e^{\frac{-t}{4}} u'' = 0\)

Answer 27

solving \(u'' = 0\) gives \(u = t\)
\(\implies \) second solution is \(u*e^{\frac{-t}{4}} = te^{\frac{-t}{4}}\)

Answer 28

@jasonjohnson86 i think that was the same potato as what rational has just said in the end

Answer 29

right....I dont think I need to run through all that for this solution though.....

Answer 30

It's just a repeated factor...not reduction of order right?

Answer 31

you dont need to, but u need to knw why you're multiplying \(t\)

Answer 32

for non-BS Math students, I personally think it's sufficient to just know that you multiply a variable of a higher degree for repeated factors =))

Answer 33

yes, if today is your last day in differential equations course