Consider the forced but undamped system described by the initial value problem $$u''+u=3\cos(\omega t)$$



(a) Find the solution $u(t)$

(b) Plot the solution u(t) versus t for $\omega = 0.7$, $\omega = 0.8$, and $\omega = 0.9$. Describe how the response $u(t)$ changes as $\omega$ varies in this interval. What happens as $\omega$ takes on values closer and closer to $1$? Note that the natural frequency of the unforced system is $\omega_0 = 1$.

(c) For the initial value problem above, plot $u'$ versus $u$ for $\omega= 0.7$, $\omega=0.8$, and $\omega= 0.9$; that is, draw the phase plot of the solution for these values of $\omega$. Use a $t$ interval that is long enough so the phase plot appears as a closed curve. Mark your curve with arrows to show the direction in which it is traversed as $t$ increases.

(a) I am able to find the solution $u(t)$ which gives $$\frac{3\cos(\omega t)-3\cos(t)}{1-\omega^2}$$

(b) I am able to plot the graph for $\omega= 0.7$, $\omega= 0.8$, and $\omega = 0.9$ which gives a sinusoidal shaped curve, I notice that for larger values of $\omega$, the amplitude of the curve become larger and its period will become smaller. Am I correct? But when I plot the graph for $\omega_0 = 1$, there was no graph to be seen, so I have to make a guess as to its properties. I mentioned that, its amplitude will tend to infinitely large while its period will tend to infinitely small. Am I correct?

(c) It is this part here that I have a problem. I tried to evaluate $u'$, plot $u'$ against $u$ parametrically but I don't seem to get the correct graph. Could anyone explain this part to me? Thanks

  • $\begingroup$ did i not answer a similar question before? $\endgroup$
    – abel
    Mar 2, 2015 at 22:30

1 Answer 1


For (a), note that your solution form is undefined for $\omega = 1$. As $\omega$ approaches $1$, the denominator would become $0$. You can see this by taking the limit $$ \lim_{\omega \to 1} \frac{3\cos(\omega t) - 3\cos t}{1-\omega^2} = 0 $$

That's the reason why your plot disappears at $\omega = 1$

The problem is when the frequency $\omega = 1$, the equation becomes

$$ u'' + u' = 3\cos t$$

You should see that the RHS looks just like the homogeneous solution, i.e. $u_p = A_1\sin t + A_2\cos t$. In order to fix this, the solution for $\omega = 1$ has to take a different form.

I'm going to guess that you used the method of undetermined coefficients, correct?

In that case, you will need to add on an additional factor of $t$ when you make your guess of the solution, like this:

$$ u(t) = C_1\,t\sin t + C_2\,t\cos t $$

Instead of what you would have guessed if $\omega \ne 1$, which is

$$ u(t) = C_1 \sin (\omega t) + C_2 \cos (\omega t)$$

I'll leave the solving to you.


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