check my work (part 2)

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

check my work (part 2)

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Question: why didn't you go directly from the original equation? Why did you go from the given solution \(\vec r(t) = \vec C e^{xt}\)
What if they don't give you the solution? how can you find it?
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 :)
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.
Oh, But I'm just starting off on this so I'm going slow :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question