Question

check my work (part 3)

Answer 1

Given:
\[\frac{d \vec Y}{dt}=\vec X\]
and\[\frac{d \vec X}{dt}=-\vec Y\]
Find
\[\vec X\]and
\[\vec Y\]
In 1st equation... integrating with respect to t
\[\vec Y=\int\limits \vec X(t)dt\]
put this in equation 2\[\frac{d \vec X}{dt}=-\int\limits \vec X(t) dt\]
diff w.r.t t\[\frac{d^2 \vec X}{dt^2}=-\vec X\]
\[\frac{d^2 \vec X}{dt^2}+\vec X=\vec 0\]
let\[\vec X(t)=\vec C e^{rt}\]
then\[\vec Cr^2e^{rt}+\vec Ce^{rt}=\vec 0\]\[\vec Ce^{rt}(r^2+1)=\vec 0\]\[\implies r^2+1=0\]\[r=\pm i\]\[\lambda=0\]\[\mu=1\]\[\vec X(t)=\vec C_{1}e^{\lambda t}\cos(\mu t)+\vec C_{2} e^{\lambda t}\sin(\mu t)\]\[\vec X(t)=\vec C_{1}\cos(t)+\vec C_{2}\sin(t)\]
\[\vec Y(t)=\int\limits \vec X(t) dt=\vec C_{1}\int\limits \cos(t)dt+\vec C_{2} \int\limits \sin(t)dt=\vec C_{1}\sin(t)-\vec C_{2}\cos(t)\]

Answer 2

It seems good to me.

Answer 3

thx