Why is $x=C\cos(\omega t)$ the solution of $m\frac{d^2 x}{dt^2}+kx=F_o\cos(\omega t)$ though lacking two arbitrary constants?

Question

I was studying undamped oscillator with harmonic driving force at the steady-state condition. It can be expressed in the form of differential equation as:$$m\dfrac{d^2 x}{dt^2}+kx=F_o\cos(\omega t).$$ Since its degree is two, the solution of this equation must have two arbitrary constants.

Now, in his book, Vibrations & waves, A.P.French deduces that $x = C\cos(\omega t)$ is the solution of the above equation as he describes as:

To obtain the steady-state solution of this equation, we set $$x = C\cos(\omega t).$$ We are assuming ,in other words, that the motion is harmonic, of the same frequency & phase as the driving force & that the natural oscillations of the system are not present. It must be kept in mind that the assumption of the solution is tentative & we must be prepared to reject it if we fail to find a value of the as-yet-undetermined constant $C$ such that The differential equation is satisfied for arbitrary values of $\omega$ & $t$. Differentiating the solution twice w.r.t., we get $$\dfrac{d^2 x}{dt^2} = -{\omega}^2 C\cos(\omega t).$$ Substituting in the differential, we thus have, $$-m{\omega}^2 C\cos(\omega t) + kC\cos(\omega t) = F_0 \cos(\omega t) \qquad \& \qquad \therefore \qquad C = \dfrac{F_0/m}{{\omega_0}^2-{\omega}^2}.$$ This satisfactorily defines $C$ in such a way that our differential equation is always satisfied. Thus, we can take it that the forced motion is indeed described by our assumed solution with $C$ depending on $\omega$ as defined above.

Here, the solution contains constant $C$ which can be readily specified by the values of $m,\omega_0,F_o,\omega$. Thus this equation contains no arbitrary constant. Then, how can $x = C\cos(\omega t)$ be the solution of the differential equation if it doesn't contain two arbitrary constants?

Answer 1

you found the particular solution to the homogoneous ODE. In particular, for the general case, if you have some $x=v(t)$ that satisfies:

$$a(t) \frac{d^{2}x}{dt^{2}} + b(t)\frac{dx}{dt} + c(t) = F(t)$$

and you have the two linearly indepdendent solutions $x_{1}(t)$ and $x_{2}(t)$ to the homogenous version of the equation:

$$a(t) \frac{d^{2}x}{dt^{2}} + b(t)\frac{dx}{dt} + c(t) = 0$$

Then, you can show that the general solution to the first equation is:

$$x(t) = v(t) + c_{1}x_{1}(t) + c_{2}x_{2}(t)$$

where the $c$'s are arbitrary constants. So, you just need to add in the two solutions to the homgenous equation.

your general solution is $x = C\cos(\omega t) + A\cos(\omega_{0}t) + B\sin(\omega_{0} t)$, which in fact does have two independent constants.

Answer 2

Because it is a solution rather than the solution. There are always infinitely many solutions to a differential equation, which we then cut down by specifying initial conditions or boundary conditions. Here, all that's been shown is that there is a $C$ such that $C \cos \omega t$ is a solution to the differential equation. That means it's possible that such motion could occur, as it satisfies the differential equation. It does not make it the unique motion of the system unless you specify further.

For instance, what if you had started with the original oscillator displaced in $x$ by some finite amount? Then you would expect that the future motion would likewise have this displacement, but there's no place for that in the solution as written.

Answer 3

You forgot about the initial conditions: position and velocity at time 0. If you add those as inputs, your solution is no longer as simple as you make it out to be. In fact - you would need two additional terms (phase, amplitude) in order to get a general solution. Note that an undamped oscillator may not have a steady state solution...