Question

Solve the following initial value problem.

\(y''-2y'+2y=0,~y(\pi)=e^{\pi},~y'(\pi)=0\)

Answer 1

\(r^2-2r+2=0\)

\(r=\dfrac{2\pm2i}{2}\)

How would you set this up?

Answer 2

r = 1\(\pm\)i

Answer 3

Sorry, I finally worked my way to the answer.

\(y(x)=e^x(-\cos(x)+\sin(x))\)

Answer 4

not - cos, just cos

Answer 5

Are you sure about that?

Answer 6

near the bottom of the first page and then continue in the second one

Answer 7

Is that your textbook?

Answer 8

hhahaha... good. I took it last summer,