Question

How would we go about solving this

Answer 1

\[m~\frac{ d^{2}x }{ dt^{2} } = -kx\]

Answer 2

where k and m are constants

Answer 3

Rewriting it like this:\[\large\rm \ddot x+\frac{k}{m}x=0\]We have a second-order linear differential equation, ya?
So let's look at the characteristic equation,\[\large\rm r^2+\frac{k}{m}=0\]
Looks like we're going to end up with sines and cosines.\[\large\rm r=\pm\sqrt{\frac{k}{m}}~i\]

Answer 4

Any confusion up to that point? :O

Answer 5

This is what my book is saying but i'm a bit confused by this: \[x = e^{-kt}+C \]

Answer 6

exponential solution? :(
Hmm that's strange... thinking...

Answer 7

Hey do you know any good videos on differential equations?

Answer 8

I can't think of too many resources off the top of my head,
but I always found MIT's OpenCourseWare to be really helpful,
even if it they are a little old :) hehe
https://www.youtube.com/watch?v=XDhJ8lVGbl8&list=PLEC88901EBADDD980
the videos on Laplace Transforms were especially useful.

Answer 9

https://www.wolframalpha.com/input/?i=mx%27%27(t)%3D-kx(t)
Wolfram seems to agree with me, sines and cosines.
I'm not sure how we're ending up with an exponential unless there are some restrictions on k like k<0 or something.

Hmm.

Answer 10

thank you