Dampened Harmonic Motion (Mathematical Modelling) 
Solve the damped harmonic motion system  $$\ddot{x} + 2k\dot{x} + \omega^2 x = 0 $$
  with initial conditions $\dot{x}= V$ at $x = 0$ in the cases
  (i) $\omega^2 = 10k^2$ (ii) $\omega^2 = k^2$ (iii) $\omega^2 = 5, k = 3$
Identify the type of damping, sketch the curve of $x$ versus $t>0$ in each case, find the zeros of $x(t)$ when applicable

$$\ddot{x} + 2k\dot{x} + \omega^2 x = 0$$
To solve this equation we can use the following
$$x = e^{\lambda t}$$ $$ \dot{x} = \lambda e^{\lambda t} $$ $$ \ddot{x} = \lambda^2 e^{\lambda t}$$
 Substitutions these terms into our original equation yields
$$\lambda^2 + 2k\lambda +\omega^2 = 0$$
This is a quadratic equation which can be solved easily 
$$\lambda = \dfrac{-2k \pm \sqrt{4k^2 - 4\omega^2}}{2}$$ $$\lambda = \dfrac{-2k \pm 2\sqrt{k^2 - \omega^2}}{2}$$ $$\lambda = -k \pm \sqrt{k^2 - \omega^2}$$
Therefore $\lambda_1 = -k + \sqrt{k^2 - \omega^2}$ and $\lambda_2 = -k - \sqrt{k^2 - \omega^2}$
We have two solutions to the quadratic equation, a linear combination of those solutions will give you the general solution. Therefore, the general solution is $$x(t) = Ae^{\lambda_1 t}+ Be^{\lambda_2 t}$$
$$x(t) = Ae^{(-k + \sqrt{k^2 - \omega^2})t} + Be^{(-k - \sqrt{k^2 - \omega^2})t}$$
(i) $\omega^2 = 10k^2$
In this case $k^2 < \omega^2$ meaning that the term in the square root will result in a negative value (two imaginary roots).
The two roots are $$\lambda_1 = -k + i\sqrt{\omega^2 -k^2 }$$  $$\lambda_2 = -k - i\sqrt{\omega^2 -k^2}$$
Again using these roots we can get a general solution
$$x(t) = Ae^{(-k + i\sqrt{\omega^2 -k^2})t} + Be^{(-k - i\sqrt{\omega^2 -k^2})t}$$
Using the relationship $$e^{i\theta} = \cos\theta + i\sin\theta $$ and $$\omega' = \sqrt{\omega^2 -k^2}$$
we get $$x(t) = Ae^{-kt}(\cos(\omega' t) + \phi)$$
This is an underdamped system since $k^2 < \omega^2$. The curve will oscillate less and less and return to equilibrium. Not too sure about when $x(t)$ will be zero. 
(ii) $ \omega^2 = k^2$ 
In this we have a real double root and just let $\omega^2 = k^2$ and substitute that into our general solution. We then use partial fractions to get the solutions $$x(t) = At + Be^{-k}$$ This system is critically damped and the curve looks to decrease exponentially over time until it reaches equilibrium. Again not too sure how to find the zeros.
(iii) $ \omega^2 = 5, k = 3$
In this case $k^2 > \omega ^2$ and we substitute in our given values to find out two roots. 
$$\lambda _1 = 1$$ $$\lambda_2 = -7$$ Using the values for  both of the roots we can find the general solution. $$ x(t) = Ae^t + Be^{-7t}$$
The type of damping in this system is over damping. The curve will  look like critical damping but it will be wont be as steep and wont reach equilibrium as fast. 
So I feel that I'm leaving out quite a bit from the questions here. For part (iii) This is the only case that gives numbers for $k$ and $\omega$, does that mean that this will be the only case where we can find zeros for $x(t)$ damping etc will look like. Or do I actually need to use numerical graphs. Also for part (iii)
 A: You are given the initial conditions $x(0)=0$ and $\dot{x}= V$. So for the first case you can write $x(t)=Ae^{-kt}\sin(\omega't)$. Then $\dot{x}=-kAe^{-kt}\sin(\omega't)+\omega'Ae^{-kt}\cos(\omega't)$. At $t=0$ we have then $V=\omega'A$, so your equation is $x(t)=\frac{V}{\omega'}e^{-kt}\sin(\omega't)$.
Similarly, you should use the initial conditions to find out the constants for the second and third case
A: Finally i think I've worked it all out. 
(ii) $k^2 = \omega^2$ the roots are degenerate. therefore $$\lambda_1 = \lambda_2 = -k$$
now the two exponents $e^{\lambda_1 t}$ = $e^{\lambda t}$ and $e^{\lambda_2 t}  = e^{\lambda t}$ are not linearly independent.
$$\ddot{x}+2k\dot{x} + k^2x =0$$ The solution to our ODE is $$x(t) = (A+Bt)e^{-kt}$$.
$$\begin{align}
    \dot{x} & = Be^{-kt} - k(A+Bt)e^{-kt} \\
    \dot{x}(0)  & = B-kA \\
            V & = B-kA
\end{align} $$   
$x(0)= A$ and 
$\dot{x}(0) = B-kA$. Using these initial conditions we find that
$$A= x(0)$$
$$B= V$$
The particular solution is $$x(t) = (Vt)e^{-kt}$$
and $x(t)$ will be zero when $t=0$.
A: (iii) $k=3, \omega^2 = 5$
In this case $k^2>\omega^2$ so we will get two real negative roots. 
$$\lambda_1 = -3 + \sqrt{3^2-5}=-1$$ $$\lambda_2 = -3 - \sqrt{3^2-5} = -5$$
Our most general solution is 
$$\begin{align}
x(t)&= Ae^{\lambda_1 t} + Be^{\lambda_2 t} \\
    &  = Ae^{-t} + Be^{-5t} \\
  \dot{x}  & = -Ae^{-t}  -5Be^{-5t} 
\end{align}
$$
Using our initial conditions we get 
$$x(0) = A+B$$ $$ \dot{x}(0) = -A -5B$$ So
$$A+B =0$$ $$-A-5B = V$$
Solving for A and B we get $B = -\frac{4}{V} $ and $A = \frac{4}{V}$
Our solution is now $$x(t)= \frac{4}{V}e^{-t}- \frac{4}{V}e^{-5t}$$
$x(t)$ will always be positive.  
