Question

\[y''-4y'+y=0\]
auxiliary equation
\[r^2-4r+1=0\]
woow I can't factor it?

Answer 1

try the quadratic formula

Answer 2

ah ok...sorry kinda rusty on algebra...hihi

Answer 3

\[r=\frac{-(-4)\pm\sqrt{(-4)^2-4\times(1)\times(1)}}{2\times(1)}\]

Answer 4

\[r=\frac{4\pm\sqrt{12}}{2}\]

Answer 5

can you simplify that square root a little bit?

Answer 6

yes sir!
\[r={2\pm\sqrt3}\]

Answer 7

brb

Answer 8

nice,
now should we check this somehow

\[\left(r-(2+\sqrt 3)\right)\left(r-(2-\sqrt 3)\right)=0\]\[r^2-r(2-\sqrt 3)-r(2+\sqrt 3)+(2+\sqrt 3)(2-\sqrt 3)=0\]\[r^2-4r+(4+2\sqrt 3-2\sqrt 3-3)=0\]\[r^2-4r+1=0\]
which is what we started with, good

Answer 9

nice! ok
so we then have>>>>>
\[y=c_1e^{(2+\sqrt3)x}+c_2e^{(2-\sqrt3)x}\]

Answer 10

\[\Large\color{red}
\checkmark\]Great work!

Answer 11

can you also write that with sines and cosines?

Answer 12

oh my

Answer 13

show me. In what way would I do that with sines and cosines?

Answer 14

Let's do that tomorrow. It's almost 1am here, and I've gotta sleep. Thanks @UnkleRhaukus !!!!

Answer 15

I was thinking of a slightly harder problem if you had\[r=2\pm\sqrt{-3}\]\[r=2\pm i\sqrt{3}\]\[y(x)=c_1e^{(2+i\sqrt3)x}+c_2e^{(2-i\sqrt3)x}\]
\[=e^{2x}\left(c_1e^{i\sqrt3}+c_2e^{-i\sqrt 3}\right)\]
now using \[e^{i\theta}=\cos\theta +i\sin\theta\]
we get
\[y(x)=e^{2x}\left(\cos \sqrt 3x+\sin \sqrt 3x\right)\],

but your question didn't have complex solutions so leave it as you had it