Question

\[y''-3y'+2y=0\]
y(0)=1 y(3)=0
auxiliary equation
\[r^2-3r+2=0\]
\[(r-2)(r-1)=0\]
\[r_1=2\]
\[r_2=1\]
\[y(0)=c_1e^{2(0)}+c_2e^0=1\]
\[y(0)=c_1+c_2=1\]
\[y(3)=c_1e^6+c_2e^3=0\]
\[c_2=\frac1{1-e^{\frac13}}\]
\[c_1=1-\frac1{1-e^{\frac13}}\]

Answer 1

e ^1/3 ?

Answer 2

e^-3 I think you mean...

Answer 3

what's wrong with it? LOL I"m not getting it? What's the difference?

Answer 4

e^1/3 is the cube root of e e^-3 is 1/(e^3)

Answer 5

oh LOL someone else was trying to explain it to me what I was too stubborn to accept it. But I guess I'll accept it now. Thanks guys and thanks @completeidiot

Answer 6

finally

Answer 7

glad you finally get it

Answer 8

yes! sigh.....thanks for being patient with me

Answer 9

no prob

Answer 10

try this what's the cube root of 8 (8^(1/3) ) ?
what's 1/(8^3) (8^-3)
they're not the same:)

Answer 11

misstype so I reposted

Answer 12

2 and 1 over 64*8

Answer 13

yep

Answer 14

very different

Answer 15

true

Answer 16

:)

Answer 17

gosh, so many new first timer questions

Answer 18

thanks again :P