Question

check my work (part 2)

Answer 1

Ok 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...
\[\vec C x^2e^{xt}+4 \vec Ce^{xt}=\vec 0\]\[\vec C e^{xt}(x^2+4)=\vec 0\]\[\implies x^2+4=0\]\[x=\pm2i\]
\[x_{1}=0+2i\]\[x_{2}=0-2i\]\[\lambda=0\]\[\mu=2\]
\[\vec r(t)=\vec C_{1}e^{\lambda t}\cos(\mu t)+\vec C_{2}e^{\lambda t}\sin(\mu t)\]\[\vec r(t)=\vec C_{1}e^0\cos(2t)+\vec C_{2}e^0\sin(2t)\]\[\vec r(t)=\vec C_{1}\cos(2t)+\vec C_{2}\sin(2t)\]

Answer 2

yup

Answer 3

thx :)

Answer 4

Question: why didn't you go directly from the original equation? Why did you go from the given solution \(\vec r(t) = \vec C e^{xt}\)

Answer 5

What if they don't give you the solution? how can you find it?

Answer 6

I 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 :)

Answer 7

From 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 2-lines solution. That is what I meant.

Answer 8

Oh, But I'm just starting off on this so I'm going slow :)