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anonymous
 one year ago
check my work (part 3)
anonymous
 one year ago
check my work (part 3)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Given: \[\frac{d \vec Y}{dt}=\vec X\] 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\[\vec Cr^2e^{rt}+\vec Ce^{rt}=\vec 0\]\[\vec Ce^{rt}(r^2+1)=\vec 0\]\[\implies r^2+1=0\]\[r=\pm i\]\[\lambda=0\]\[\mu=1\]\[\vec X(t)=\vec C_{1}e^{\lambda t}\cos(\mu t)+\vec C_{2} e^{\lambda t}\sin(\mu t)\]\[\vec X(t)=\vec C_{1}\cos(t)+\vec C_{2}\sin(t)\] \[\vec Y(t)=\int\limits \vec X(t) dt=\vec C_{1}\int\limits \cos(t)dt+\vec C_{2} \int\limits \sin(t)dt=\vec C_{1}\sin(t)\vec C_{2}\cos(t)\]
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