Solving a second order inhomogeneous differential equation with constant coeffcients I a seeking to solve the equation: $$u'' + 2vu' + u = cos(\sigma t), \ u(0) = 1, \ u'(0) =0$$ where $0 < v < 1$. Then I have to show that the solutio is purely oscillatory (which I don't know how is defined) and find the amplitude of the solution as a function of $\sigma$.
Trying to find a particular integral ended up being horribly messy. If anyone could help it will be deeply appreciated.
 A: First find the homogeneous solution: 
$$u(t) = e^{\alpha t}$$
And plug it in to the homogeneous equation to find $\alpha$:
$$\alpha^2+2v\alpha+1=0\ \to\ \alpha =\pm\sqrt{v^2-1}-v$$
Now it remains to find (guess) a non-homogeneous solution. Assume:
$$u(t) = A\cos(\sigma t) + B\sin(\sigma t)$$
Then:
$$-\sigma^2 (A\cos(\sigma t) + B\sin(\sigma t)) - 2v\sigma (A\sin(\sigma t) - B\cos(\sigma t)) + A\cos(\sigma t) + B\sin(\sigma t) =\cos(\sigma t)$$
You'll find that:
$$A-2Bv\sigma -A\sigma^2 = 0$$
$$B+2Bv\sigma -B\sigma^2 = 1$$
And from here you can solve for $A,B$. The final solution is given by the sum:
$$u(t) = C_1e^{(\sqrt{v^2-1}-v)t} + C_2e^{(-\sqrt{v^2-1}-v)t} + A\cos(\sigma t) + B\sin(\sigma t)$$
Now use the initial conditions:
$$u(0)=1 \ \to \ C_1+C_2 +A=1$$
$$u'(0)=0 \ \to \ C_1(\sqrt{v^2-1}-v)+C_2(-\sqrt{v^2-1}-v) +B=0$$
And thus find out the constants $C_1$ and $C_2$. From the looks of it, this is not a purely oscillatory solution, since $C_1$ and $C_2$ are not zero (at least not for all values of $v$. 
A: The solution is split into a homogeneous $u^{(H)}$ solution and inhomogeneous $u^{(I)}$ solution. $u^{(H)}$ satisfies
$$u^{(H)''}+ 2 v u^{(H)'} + u^{(H)}= 0$$
Assume $u^{(H)} = A e^{r t}$.  Then $r$ satisfies $r^2 + 2 v r + 1=0 \implies r = -v \pm i \sqrt{1-v^2}$ since $0<v<1$.  We may then write
$$u^{(H)}(t) = e^{-v t}[ A e^{i \sqrt{1-v^2} t} + B e^{-i \sqrt{1-v^2} t}]$$
For $u^{(I)}$, guess that $u^{(I)}(t) = P \cos{(\sigma t)} + Q \sin{(\sigma t)}$.  Then
$$ \cos{(\sigma t)} [(1-\sigma^2) P + \sigma Q] + \sin{(\sigma t)} [(1-\sigma^2) Q - \sigma P] = \cos{(\sigma t)}$$
Equating coefficients, we get
$$\begin{align}(1-\sigma^2) P + \sigma Q &=1 \\ - \sigma P+ (1-\sigma^2) Q &=0\\ \end{align}$$
which means that
$$P = \frac{1-\sigma^2}{(1-\sigma^2)^2 + \sigma^2}$$
$$Q = -\frac{\sigma}{(1-\sigma^2)^2 + \sigma^2}$$
$A$ and $B$ are determined by the initial conditions you specified:
$$u^{(H)}(0) = 1 \implies A + B + P =1 $$
$$u^{(H)'}(0) = 0 \implies ( -v + i \sqrt{1-v^2}) A + ( -v - i \sqrt{1-v^2}) B + \sigma Q =0 $$
Then
$$A = (\frac{1}{2} (1-P) - \sigma Q) (1+v + i \sqrt{1-v^2}) $$
$$B = \frac{1}{2} (1-P)(1-v - i \sqrt{1-v^2}) + \sigma Q (1+v + i \sqrt{1-v^2})$$
The net solution is then
$$u(t) = u^{(H)}(t) + u^{(I)}(t) = e^{-v t}[ A e^{i \sqrt{1-v^2} t} + B e^{-i \sqrt{1-v^2} t}] + P \cos{(\sigma t)} + Q \sin{(\sigma t)} $$
Note the presence of the damping term implies non-pure oscillation.
