Is a constant on the RHS of the equation of simple harmonic motion allowed? I read at a STEP booklet that we have to know how to bring a simple harmonic motion's equation to the form: 
$$\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + \omega^2x= c$$
where $c$ is a constant. We also have to quote the solution.
But shouldn't $c$ be zero always? (since $\omega^2=\frac{k}{m}$ and $k\cdot\text{displacement}/m = \text{acceleration}$)
Could you also tell me what the solution I have to quote can be?
 A: Let me say basically the same thing as Rob, but in what seems to me a simpler way.
If you start with the equation:
$$ \frac{\mathrm{d}^2x}{\mathrm{d}t^2} + \omega^2x= c \tag{1} $$
We can rearrange it as:
$$ \frac{\mathrm{d}^2x}{\mathrm{d}t^2} + \omega^2\left(x- \frac{c}{\omega^2}\right) = 0 $$
Then you can define a new variable $z$ by:
$$ z = x - \frac{c}{\omega^2} $$
The constant $c/\omega^2$ disappears when you differentiate, so using the new variable $z$ our equation (1) becomes:
$$ \frac{\mathrm{d}^2z}{\mathrm{d}t^2} + \omega^2z= 0 $$
So your constant $c$ can be eliminated with a simple change of origin i.e. moving the origin by a distance $c/\omega^2$.
A: The equation
$$ \frac{d^2 x}{dt^2} + \omega^2 x = 0$$
is an example of a homogeneous second order linear differential equation, with a general solution of the form
$$x = A\sin(\omega t) + B\cos(\omega t),$$
where $A$ and $B$ are constants to be determined from the boundary conditions of the problem.
If you now put a constant on the right hand side of ths equation, it becomes an inhomogeneous second order differential equation.
The general solution to this will be the sum of the general solution you found for the homogeneous equation plus something that is normally called the particular solution $P(t)$.
i.e.
$$ x(t) = A\sin(\omega t) + B\cos(\omega t) + P(t)$$
Now because we know that substituting the first two terms into the left hand side of the differential equation will give zero (because they are the general solution to the homogeneous equation) then we can also say that
$$ \frac{d^2 P}{dt^2} + \omega^2 P = c$$
To be true for all values of $t$, we see that $P$ must in fact also be a constant, so $P = c/\omega^2$.
The final general solution would be
$$ x(t) = A\sin(\omega t) + B\cos(\omega t) + \frac{c}{\omega^2}$$
Or: (as JR has pointed out)
$$ x(t) - \frac{c}{\omega^2} = A\sin(\omega t) + B\cos(\omega t)$$
You can confirm this works by substituting back into the original differential equation.
The method I've used above is also well-suited to more complicated functions of time on the RHS, using an appropriate guess at the form of $P(t)$.
