Deriving the particular solution for a damped driven harmonic oscillator Consider a damped driven harmonic oscillator, for which $\beta = \omega_0/4$ and the driving force is given by $F = F_0\cos\omega t$ ($\omega_0$ and $F_0$ represent initial condition of those variables). Our goal is to find the general solution, which is the sum of the complementary and particular solutions:
The differential equation is the following:
$$\ddot{x} + 2\beta\dot x + \omega_0^2x = F_0\cos\omega t$$
c) For the particular solution, try the following function:
$$x_p(t) = B_1\cos\omega t + B_2\sin\omega t = F_0\cos\omega t$$
Substitute the trial solution into your equation of motion, rearrange, and equate the coefficients to find expressions for $B_1$ and $B_2$. 
First, the first derivative and second derivatives of the particular solution were taken, which is the following:
\begin{align}
x_p(t) &= B_1\cos\omega t + B_2\sin\omega t\\
\dot x_p(t) &= -\omega B_1\sin\omega t + \omega B_2\cos\omega t\\
\ddot x_p(t) &= -(\omega^2)B_1\cos\omega t - \omega^2B_2\sin\omega t
\end{align}
Plugging into the differential equation we get the following:
$$(-\omega^2B_1\cos\omega t - \omega^2B_2\sin\omega t) + 2\beta\omega(-B_1\sin\omega t + B_2\cos\omega t) + \omega_0^2(B_1\cos\omega t + B_2\sin\omega t) = F_0\cos\omega t$$
Then the terms were simplified into cos and sin:
$$[-\omega^2B_1 + 2\beta\omega B_2 + \omega_0^2B_1]\cos\omega t + [-\omega^2B_2 - 2\beta\omega B_1  + \omega_0^2B_2]\sin\omega t = F_0\cos\omega t$$
How do I obtain the equations for B(1) and B(2) when there is only one equation present? 
Using G. Paily's suggestion, since there does not exist any constant such that csinωt = F(0)cosωt, then the sin terms must be equal to zero. Using this fact, we can create two equations: cos terms = F(o)cosωt and sin terms = 0. First use the sin terms to figure out what B(2) is, which is the following:
B(2) = 2βωB(1)/(ω(o)^2 -ω^2)
Taking B(2) and substituting into the cos terms after simplification, I we are left with:
B(1)cosωt(ω(o)^2 + 4(β^2)(ω^2)/(ω(o)^2-ω^2) - ω^2) = F(o)cosωt
After substituting for β and simplifying again, we derive the equation for B(1):
B(1) = F(o)/(ω(o)^2 + ω^2/[4(ω(o)^2-1)] -ω^2 
The final answer is the following:
B(1) = F(o)/(ω(o)^2 + ω^2/[4(ω(o)^2-1)] -ω^2
B(2) = 2βωB(1)/(ω(o)^2 -ω^2)
Thank you for your help.
 A: The fact that you can eliminate the $\sin$ term tells you that its coefficient $[-\omega^2 B_{2}-2\beta\omega B_{1}+\omega_{0}^{2}B_{2}]$ must be  zero. This gives you another relation between $B_{1}$ and $B_{2}$, and with two equations and two unknowns, you can solve cleanly for $B_{1}$ and $B_{2}$.
A: First of all you have a mistake in calculus: you say "Plugging into the differential equation I get: ..." Do check, you improperly substituted x¨_p (t). You left a minus inside the round parentheses that should be plus. So, in fact you have, not minus.
−ω^2 B_1 cos(ωt)     − ω^2 B_2 sin(ωt)
+2βω B_2 cos(ωt)     − 2βω B_1 sin(ωt) 
+(ω_0)^2 B_1 cos(ωt) + (ω_0)^2 B_2 sin(ωt)
-F_0 cos(ωt)         = 0 
So, you can immediately get your equations for B_1 and B_2, by equating separately the coefficients of B_1, and those of B_2 ,
[(ω_0)^2 − ω^2]B_1 + 2βω B_2 = F_0,
[(ω_0)^2 − ω^2]B_2 − 2βω B_1 = 0 .
Now, from the 2nd equality you get, 
(1) B_2 = 2βω B_1/[(ω_0)^2 − ω^2] ,
and substituting in the 1st equation you get an equation exclusively for B_1:
B_1{[(ω_0)^2 − ω^2] + (2βω)^2 /[(ω_0)^2 − ω^2]} = F_0 ,
from which
(2) B_1 = F_0 / {[(ω_0)^2 − ω^2] + (2βω)^2 /[(ω_0)^2 − ω^2]} ,
Substituting in (1),
(3) B_2 = 2βω F_0/{[(ω_0)^2 − ω^2]^2 + (2βω)^2} .
You see, the interesting conclusion is that for ω = ω_0 you have an amplification of the sinusoidal part of the solution, B_2 = F_0/(2βω) , while the cosine part becomes zero (B_1 = 0). On the other hand, if β=0, the wave remains purely cosine, as the driving force, and what is more, if ω = ω_0, B_1 -> infinity.
Good luck ! 
