# 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?

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?

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.

• Oh! he is discussing only a part of the whole story! But just think, if you are given a differential equation of 2nd degree, what solution do you assume? Wouldn't it be a function having two arbitrary constants? Then why did Mr.French assume such a solution at first which contained only one arbitrary constant $C$ (though later it was proved that this solution doesn't contain any arbitrary constant!)? Mean to say why did he approach the problem in such a way instead of taking a function having two parameters? I was taught to solve such equations assuming a function having two parameters. – user36790 Jun 5 '15 at 21:16
• @user36790: he solved the interesting part of the problem. All that those $A$ and $B$ factors are good for is setting your solution up to match initial conditions. There is always a case where $A = B = 0$, for some initial values of $x$ and $v$. – Jerry Schirmer Jun 5 '15 at 21:59
• Also, I would say that you don't assume an answer in either case. you guess and answer and then you check it. – Jerry Schirmer Jun 5 '15 at 22:59
• It should also be noted that, for $\omega = \pm \omega_{0}$, your solution is singular, and you will have to rework the whole thing from scratch. – Jerry Schirmer Jun 5 '15 at 23:03

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.

• Also, the quote in the OP is looking for the steady state solution, which is unique. The arbitrary constants appear in the transient solution. – Javier Jun 5 '15 at 20:22
• Oh, I totally missed that. Thanks. (Though I guess now you'd want to prove that the steady state solution really is unique.) – zeldredge Jun 5 '15 at 20:23
• Hang on. This is an undamped oscillator; the words "steady state" are misleading, since the "transient" won't decay with time. You could show that every solution is a sum of oscillations with frequencies $\sqrt{k/m}$ and $\omega$, and the derivation in the OP shows that there is no freedom left in the "steady state". – Javier Jun 5 '15 at 20:26
• Well, the OP derivation also mentions "the natural oscillations of the system are not present." So I think they assume that there are no $\omega_0$ components present. Since those satisfy the homogeneous equation, they'll always give zero on the LHS of the equation. There is freedom, since you can always add a homogeneous solution, but we're just not looking at that case. – zeldredge Jun 5 '15 at 20:33

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...

• Well, sir, I've done nothing at all; all I wanted is to comprehend the logic of assumption of a solution of second-degree differential equation that was made by Mr.French in his deduction. – user36790 Jun 5 '15 at 20:41
• OK - Mr French omitted talking about initial conditions... Given the considerable contributions he has made, I forgive him :-) – Floris Jun 5 '15 at 20:49
• After much reading, I infer that Mr. French is dealing with only a part of the whole story: the other piece contains the desired two arbitrary constants. +1 :) – user36790 Jun 5 '15 at 21:02
• Can you tell me, sir, why Mr. French used such an approach to solve the equation? Because if you are given a differential equation of 2nd order, you must assume a function having two parameters unlike Mr.French who used such a function which has no parameter at all! – user36790 Jun 5 '15 at 21:22
• Not at all. You must not anything. It's simply what you are taught in calculus, but everyone's welcome to take shortcuts when they know what they're doing. Say, I had a driven damped harmonic oscillator, it would surely make no sense for me to consider the solution of the homogeneous equation (the one with two arbitrary parameters) if all I cared about was the steady state, as the natural oscillations would quickly decay to zero. – LLlAMnYP Jun 6 '15 at 0:19