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anonymous
 one year ago
check my work (part 2)
anonymous
 one year ago
check my work (part 2)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so this time I've studied more pls check my work:) \[\frac{d^2 \vec r}{dt^2}+4\vec r=\vec 0\] Let\[\vec r(t)=\vec Ce^{xt}\] The equation is then... \[\vec C x^2e^{xt}+4 \vec Ce^{xt}=\vec 0\]\[\vec C e^{xt}(x^2+4)=\vec 0\]\[\implies x^2+4=0\]\[x=\pm2i\] \[x_{1}=0+2i\]\[x_{2}=02i\]\[\lambda=0\]\[\mu=2\] \[\vec r(t)=\vec C_{1}e^{\lambda t}\cos(\mu t)+\vec C_{2}e^{\lambda t}\sin(\mu t)\]\[\vec r(t)=\vec C_{1}e^0\cos(2t)+\vec C_{2}e^0\sin(2t)\]\[\vec r(t)=\vec C_{1}\cos(2t)+\vec C_{2}\sin(2t)\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Question: why didn't you go directly from the original equation? Why did you go from the given solution \(\vec r(t) = \vec C e^{xt}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1What if they don't give you the solution? how can you find it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't understand what you mean ?? you just solve the polynomial and make the equations accordingly if variable is real and distinct, real and same and complex right because IT IS the method for solving the equation, I just didn't take exponential function out of the blue :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1From r" +4r =0 characteristic equation is \(\lambda ^2 +4 =0\) which gives us \(\lambda = \pm 2i\) hence the general solution is as you get \(r = C_1cos(2t) + C_2 sin(2t)\) it is kind of 2lines solution. That is what I meant.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, But I'm just starting off on this so I'm going slow :)
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