MX + KX = 0 (I NEED YOUR HELP)

[eduardoferrer2k]

I need found the general solve to the next diferential equiations:

mx'' + kx = 0 (a spring without damper)
mx'' + dx' + kx' = 0 (a spring with damper)

thank you

 

[daon]

Are m, d and k constants?

 

[eduardoferrer2k]

yes, are constants

 

[daon]

I take it you are in an ODE class?

1. mx'' + kx = 0 (a spring without damper)

\(\displaystyle mr^2 + k = 0\)

\(\displaystyle r = \sqrt{\frac{-k}{m}\)

Since there is only one root, so the solution is given by:

\(\displaystyle \L\\ c_1e^{r_1x} + c_2xe^{r_1x}\) = \(\displaystyle \L\\ c_1e^{\frac{-kx}{m}} + c_2xe^{\frac{-kx}{m}}}\)

For the second problem, start by factoring out the x':

\(\displaystyle \L\\ mx'' + (d+x)x'= 0\)

Apply the same principal.

\(\displaystyle \L\\ mr^2 + (d+x)r= 0\)

In this case you will have 2 real roots, so the solution is given by:

\(\displaystyle \L\\ c_1e^{r_1x} + c_2e^{r_2x}\)

Check my work, please.