1. Your solution will be $x(t) = C_1 \cos w_1 t + C_2 \sin w_1 t$. You just set $w=w_1$. You could very well have set $w=w_2$ this would just change the values of $C_1,C_2$. The conventions is to set $w=w_1$, but it doesn't matter.

2. Originally Posted by lvleph
Your solution will be $x(t) = C_1 \cos w_1 t + C_2 \sin w_1 t$. You just set $w=w_1$. You could very well have set $w=w_2$ this would just change the values of $C_1,C_2$. The conventions is to set $w=w_1$, but it doesn't matter.

The solution is:
$x(t) = e^{-(\frac{c}{2m})t}(C_1coswt+C_2sinwt)$

I do not understand why the $e^{-(\frac{c}{2m})t}$ ???

3. I was just about to write this. I made a mistake. Say $w$ is of the form $w = a + bi$ then the solution should be of the form
$x(t) = C_1 e^{at} \cos bt + C_2 e^{at} \sin bt$.
So, in your case above you should actually have the solution
$x(t) = C_1 e^{-\frac{c}{2m}t} \cos \frac{\sqrt{c^2 - 4mc}}{2m} t + C_2 e^{-\frac{c}{2m}t} \sin \frac{\sqrt{c^2 - 4mc}}{2m} t$.
I apologize for the oversight.

4. Originally Posted by lvleph
I was just about to write this. I made a mistake. Say $w$ is of the form $w = a + bi$ then the solution should be of the form
$x(t) = C_1 e^{at} \cos bt + C_2 e^{at} \sin bt$.
So, in your case above you should actually have the solution
$x(t) = C_1 e^{-\frac{c}{2m}t} \cos \frac{\sqrt{c^2 - 4mc}}{2m} t + C_2 e^{-\frac{c}{2m}t} \sin \frac{\sqrt{c^2 - 4mc}}{2m} t$.
I apologize for the oversight.

Why is the equation of the form?

$x(t) = C_1 e^{at} \cos bt + C_2 e^{at} \sin bt$

and why $w = \frac{\sqrt{4mk-c^2}}{2m}$ ?

5. Like I said I made a mistake and $w = w_1$. Previously, I assumed that the eigenvalues were purely imaginary which leads to a different solutions. This was incorrect. The form of the solution comes from Euler's Formula, i.e.,
$e^{a \pm i b} = e^a \cos b \pm e^a i \sin b$.
So if our solution has complex eigenvalues $w_1,w_2$ the solution
$x(t) = C_1 e^{w_1t} + C_2 e^{w_2t}$
where $w_1 = a+bi,\; w_2=a-bi$ becomes
$x(t) = C_1\left(e^{at} \cos bt + e^{at} i \sin bt\right) + C_2\left(e^{at}\cos bt - e^{at} i \sin bt\right)$.
After further simplification the solution becomes
$x(t) = C_1 e^{at} \cos bt + C_2 e^{at} \sin bt$.
Do you see why my previous solution works if the eigenvalues are purely imaginary?

Again, I am sorry for the confusion.

6. Originally Posted by lvleph
Like I said I made a mistake and $w = w_1$. Previously, I assumed that the eigenvalues were purely imaginary which leads to a different solutions. This was incorrect. The form of the solution comes from Euler's Formula, i.e.,
$e^{a \pm i b} = e^a \cos b \pm e^a i \sin b$.
So if our solution has complex eigenvalues $w_1,w_2$ the solution
$x(t) = C_1 e^{w_1t} + C_2 e^{w_2t}$
where $w_1 = a+bi,\; w_2=a-bi$ becomes
$x(t) = C_1\left(e^{at} \cos bt + e^{at} i \sin bt\right) + C_2\left(e^{at}\cos bt - e^{at} i \sin bt\right)$.
After further simplification the solution becomes
$x(t) = C_1 e^{at} \cos bt + C_2 e^{at} \sin bt$.
Do you see why my previous solution works if the eigenvalues are purely imaginary?

Again, I am sorry for the confusion.

Thank you. I still have a doubt, why $w_1 = \frac{-c}{2m} + wi$ and $w_2 = \frac{-c}{2m} - wi$

I do not understand these two equations

7. Sorry there is still confusion. Those formulations for $w_1,w_2$ are not correct. The correct one are the solutions to $mw^2 + cw + k=0$.

8. Originally Posted by lvleph
Sorry there is still confusion. Those formulations for $w_1,w_2$ are not correct. The correct one are the solutions to $mw^2 + cw + k=0$.
This equation is correct
$x(t) = e^{-(\frac{c}{2m})t}(C_1coswt+C_2sinwt)$

9. The correct solution is
$
x(t) = C_1 e^{-\frac{c}{2m}t} \cos \frac{\sqrt{c^2 - 4mc}}{2m} t + C_2 e^{-\frac{c}{2m}t} \sin \frac{\sqrt{c^2 - 4mc}}{2m} t
$
.

10. Originally Posted by lvleph
The correct solution is
$
x(t) = C_1 e^{-\frac{c}{2m}t} \cos \frac{\sqrt{c^2 - 4mc}}{2m} t + C_2 e^{-\frac{c}{2m}t} \sin \frac{\sqrt{c^2 - 4mc}}{2m} t
$
.

What is: $\frac{\sqrt{c^2 - 4mc}}{2m} t$ ???

11. $\frac{\sqrt{c^2 - 4mk}}{2m} t$ is the imaginary portion of $w_1\cdot t$ and $\frac{-c}{2m} t$ is the real part of $w_1\cdot t,w_2\cdot t$.

12. Originally Posted by lvleph
$\frac{\sqrt{c^2 - 4mc}}{2m} t$ is the imaginary portion of $w_1\cdot t$ and $\frac{-c}{2m} t$ is the real part of $w_1\cdot t,w_2\cdot t$.

Thank you very much

13. The imaginary part is not $\frac{\sqrt{c^2-4mk}}{2m}$ ???

14. Sorry I had a typo.

15. Originally Posted by lvleph