Question

Pls. Help me answer this.

y”+4y’+4y=0 ; y(0)=1, y’(0)=1

Answer 1

where are you stuck?

Answer 2

y”+4y’+4y=0 ; y(0)=1, y’(0)=1
r^2 +4r+4=(r+2)(r+2)
r=-2 and -2

Answer 3

y=c1e^-2t +C2e^-2t
y=(0)=1 y'(0)=1
y'=-C1e^-2t-2C2e-2t

Answer 4

is it correct???

Answer 5

hold on, y seems not good to me
y = e^(-2t) (C1+C2t)

Answer 6

double root r =-2 give you 2 solutions: \(e^{-2t} ~~and~~ t*e^{-2t}\)

Answer 7

general solution is
\[y = e^{-2t}(C_1 +C_2t)\]

Answer 8

got me?

Answer 9

sorry I'm confused. can you pls. explain.

Answer 10

where did i got wrong?

Answer 11

OK, to yours, C1 e^(-2t) and C2 e^(-2t) is not correct. Re read the material. To DOUBLE ROOT. the second solution MUST time with t

Answer 12

your characteristic equation gives you r = -2 twice, right?

Answer 13

yes

Answer 14

therefore, the solution are:
x1 = C1e^(-2t) ok?

Answer 15

ok

Answer 16

X2 = C2 *t *e^(-2t)
that's the formula,

Answer 17

because if you don't time t, so, x1 = x2? when by definition, the solutions must be linear INdependent. to have it, you must time t to x1 to get x2

Answer 18

so instead of y'=-C1e^-2t-2C2e-2t
it should be like this -C1e^-2t- C2 *t *e^(-2t)?

Answer 19

\[y = C_1e^{-2t}+C_2te^{-2t}\]
take derivative, you get what? do it carefully, please

Answer 20

\[y' = -2C_1e^{-2t} +C_2e^{-2t} -2C_2e^{-2t}\] got this part?

Answer 21

\(C_2 te^{-2t}\) take product rule for t and e^(-2t)

Answer 22

yes i got that part

Answer 23

so, you have both y and y'
and then, replace initial condition on them to get C1, C2

Answer 24

got it?