Question

Can anyone sole the IVP y'' + 2ky' + ( (k^2) + (w^2))y = 0, y(0) = 1, y'(0) = -k

Answer 1

Wow, that's a hairy one. Are k and w constants?

Answer 2

usually with this sort of IVP you'd use laplace iirc

Answer 3

yes they are constants

Answer 4

you hev to use homogeneous linear ODEs to solve it

Answer 5

step 1: set up characteristic equation
\[r^{2} +2kr + (k^{2}+w^{2})=0\]
\[r = -k \pm wi\]
complex roots, so use Eulers Formula to put in sin/cos form
general solution
\[y(t) = c_{1}e^{-kt}\cos(wt) + c_{2}e^{-kt}\sin(wt)\]
apply initial conditions
y(0) = 1 --> c1 = 1

\[y'(t) = kwe^{-kt}\sin(wt) - kwc_{2}e^{-kt}\cos(wt)\]
y'(0) = -k
--> -kwc2 = -k
--> c2 = 1/w
\[y(t) = e^{-kt} \cos(wt) + \frac{1}{w}e^{-kt} \sin(wt)\]