Steady periodic solution to $x''+2x'+4x=9\sin(t)$ Find the steady periodic solution to the differential equation
$x''+2x'+4x=9\sin(t)$
in the form
$x_{sp}(t)=C\cos(\omega t−\alpha)$, with $C > 0$ and $0\le\alpha<2\pi$.
I don't know how to begin. First of all, what is a steady periodic solution? And how would I begin solving this problem?
 A: To a differential equation you have two types of solutions to consider: homogeneous and inhomogeneous solutions.
The first is the solution to the equation
$$x''+2x'+4x=0$$
Taking the tried and true approach of method of characteristics then assuming that $x~e^{rt}$ we have:
$$r^2+2r+4=0 \rightarrow (r-r_-)(r-r+)=0 \rightarrow r=r_{\pm}$$
$$r_{\pm}=\frac{-2 \pm \sqrt{4-16}}{2}= -1\pm i \sqrt{3}$$ 
We see that the homogeneous solution then has the form of decaying periodic functions:
$$x_{homogeneous}= Ae^{(-1+ i \sqrt{3})t}+ Be^{(-1- i \sqrt{3})t}=(Ae^{i \sqrt{3}t}+ Be^{- i \sqrt{3}t})e^{-t}$$
Again, these are periodic since we have $e^{i\omega t}$, but they are not steady state solutions as they decay proportional to $e^{-t}$.
The other part of the solution to this equation is then the solution that satisfies the original equation:
$$D[x_{inhomogeneous}]= f(t)$$ 
Upon inspection you can say that this solution must take the form of $Acos(\omega t) + Bsin(\omega t)$. That is because the RHS, f(t), is of the form $sin(\omega t)$.  You then need to plug in your expected solution and equate terms in order to determine an appropriate A and B. Once you do this you can then use trig identities to re-write these in terms of c, $\omega$, and $\alpha$.
A: Since the real parts of the roots of the characteristic equation is $-1$, which is negative, as $t \to \infty$, the homogenious solution will vanish. We only have the particular solution in our hands.
$$\eqalign{x_p(t) &= A\sin(t) + B\cos(t)\cr
x_p'(t) &= A\cos(t) - B\sin(t)\cr
x_p''(t) &= -A\sin(t) - B\cos(t)\cr}$$
Then the equation will be
$$(-A - 2B + 4A)\sin(t) + (-B + 2A + 4B)\cos(t) = 9\sin(t)$$
$$\eqalign{3A - 2B &= 1\cr
2A + 3B &= 0\cr}$$
$$A = 3/13, B = -2/13$$
Therefore steady state solution is $\displaystyle x_p(t) = \frac{3}{13}\,\sin(t) - \frac{2}{13}\,\cos(t)$
A: 
The steady periodic solution is the particular solution of a differential equation with damping. 

Solution: Given differential equation is$$x''+2x'+4x=9\sin t \tag1$$
First, the form of the complementary solution must be determined in order to make sure that the particular solution does not have duplicate terms. 
Roots of the trial solution is $$r=\frac{-2\pm\sqrt{4-16}}{2}=-1\pm i\sqrt3$$
The general form of the complementary solution  (or transient solution) is $$x_{c}=e^{-t}\left(a~\cos(\sqrt 3~t)+b~\sin(\sqrt 3~t)\right)$$where $~a,~b~$ are constants.
Let us assumed that the particular solution, or steady periodic solution is of the form $$x_{sp} =A \cos t + B \sin t$$
Here our assumption is fine as no terms are repeated in the complementary solution. The
general form of the particular solution is now substituted into the differential equation  $(1)$ to determine the constants $~A~$ and $~B~$.
We have $$(-A\cos t -B\sin t)+2(-A\sin t+B\cos t)+4(A \cos t + B \sin t)=9\sin t$$
$$\implies (3A+2B)\cos t+(-2A+3B)\sin t=9\sin t$$
Comparing we have $$A=-\frac{18}{13},~~~~B=\frac{27}{13}$$ 
So the steady periodic solution is $$x_{sp}=-\frac{18}{13}\cos t+\frac{27}{13}\sin t$$
This particular solution can be converted into the form $$x_{sp}(t)=C\cos(\omega t−\alpha)$$where $\quad C=\sqrt{A^2+B^2}=\frac{9}{\sqrt{13}},~~\alpha=\tan^{-1}\left(\frac{B}{A}\right)=-\tan^{-1}\left(\frac{3}{2}\right)=-0.982793723~ rad,~~ \omega= 1$
The value of $~\alpha~$ is in the $~4^{th}~$ quadrant. Since $~B~$ is
positive and $~A~$ is negative, $~α~$ must be in the $~3^{rd}~$ quadrant. So $~α = π -0.982793723 = 2.15879893059 ~$. 
So the steady periodic solution is $$x_{sp}(t)=\left(\frac{9}{\sqrt{13}}\right)\cos(t−2.15879893059)$$

The general solution is $$x(t)=e^{-t}\left(a~\cos(\sqrt 3~t)+b~\sin(\sqrt 3~t)\right)+\frac{1}{13}(-18 \cos t + 27 \sin t)$$
