Question

How did I do? Second Order Differential equations
\[y'' - 8y' + 12y = 0\]
the auxiliary equation is
\[r^2-8r+12=0\]
\[(r-6)(r-2)=0\]
so r=6 and r=2
\[y=c_1e^{6x}+c_2e^{2x}\]

Answer 1

those seem right o.O

Answer 2

To check whether you have the right solution, try putting the expression obtained for y into the original differential equation. It should give you 0 = 0, if its correct.

Answer 3

actually..you're supposed to get the first derivatie of that and the second then substitute accordingly...

Answer 4

would I need to find the values of c_1 and c_2?

Answer 5

nope. you're not given any condiitons

Answer 6

you can only get the G.S.

Answer 7

Aaaaah. I seeeeeee....
\[y'=6c_1e^{5x}+c_2e^{x}\]
\[y''=30c_1e^{4x}+c_2e^{x}\]
let me double check using wolfram...I'm sure I have errors lol

Answer 8

The auxiliary equation doesn't lie. I don't see what's the problem here.

Answer 9

oh so...this should be sufficient
\[y=c_1e^{6x}+c_2e^{2x}\]

Answer 10

yes, unless you were given initial conditions.

Answer 11

\[y = c_1 e^{6x} + c_2 e^{2x}\]
\[y ^\prime = 6c_1 e^{6x} + 2c_2 e^{2x}\]
\[y^{\prime \prime} = 36 c_1 e^{6x} + 4c_2 e^{2x}\]
\[\implies 36c_1 e^{6x} + 4c_2 e^{2x} - 8(6c_1 e^{6x} + 2c_2 e^{2x}) + 12(c_1 e^{6x} + c_2 e^{2x})\]
\[\implies 36c_1 e^{6x} + 4c_2 e^{2x} - 48c_1 e^{6x} -16c_2 e^{2x} + 12c_1 e^{6x} + 12c_2 e^{2x})\]
wonder if that comes out as 0

Answer 12

yep you're safe. that's 0

Answer 13

without y'(0) and y(0) you can't find \(c_1\) and \(c_2\)

Answer 14

if there was a condition...

Answer 15

technically it can be y(anything) and y'(anything)

Answer 16

you can find it

Answer 17

btw @MathSofiya you might want to review derivatives... derivative of \[e^{6x}\] is not \[6e^{5x}\]
:S

Answer 18

good good, that makes more sense now.
Thanks everyone!!!!!....and thanks for the review on how to do derivatives of e...I did it wrong :P

Answer 19

lol welcome

Answer 20

for both