How to verify $y_c(x)+y_p(x)$ is a solution to the differential equation? I am given a nonhomogeneous differential equation:

$$y''+4y'+3y=g(x)$$ 

where $g(x)=3 \sin 2x$.
After working through the problem, I have
$$y_c(x)=C_1e^{-3x}+C_2e^{-x}$$ 
(I was to find a general solution for which $g(x)=0$) 
$$y_p(x)=-(24/65) \cos 2x-(3/65) \sin 2x. $$ 
(On this part, I was given $y_p(x)=A \cos 2x+ B \sin 2x$)
Now I'm stuck.  How do I verify that $y_c(x)+y_p(x)$ is a solution to the differential equation?
Thanks!

Well, your solution $y_{c}(x)$ satisfies the problem $y'' + 4y' + 3y = 0$ and $y_{p}(x)$ satisfies the problem $y'' + 4y'+ 3y = g(x)$. So, $(y_{c}+y_{p})'' + 4(y_{c}+y_{p})' + 3(y_{c}+y_{p}) = [y_{c}'' + 4y_{c}' + 3y_{c}] + [y_{p}'' + 4y_{p}' + 3y_{p}] = 0 + g(x) = g(x)$. Hence, $y(x)$ satisfies the ODE. Note that derivatives of any order are additive; that is $(y_{1}+y_{2})^{(n)}(x) = y_{1}^{(n)}(x) + y_{2}^{(n)}(x)$.
