# 2nd order linear ODE with constant coefficients For question $3$, I know d) is not an option since the period is not the same.

For question 4, I know that discriminant must be negative, so the coefficient of $y$ is $1$, but I'm not sure about the right hand side.

Can someone please help, thanks!

• For 3, note that you should be able to form two linearly independent solutions in a SOLDE... (second order linear differential equation) – pie314271 Mar 7 '17 at 20:02
• @pie314271 Why not all 3 of them be linearly dependent? it is possible. – Little Rookie Mar 7 '17 at 20:06
• For 3 there's a lot that can go wrong. For instance look at the third one: can a homogeneous ODE of this type have sinusoidal solutions that only differ by a constant? That would mean that constants themselves are homogeneous solutions. For 4 you're seeing resonance: the amplitude grows with time. That means that the frequency of the right hand side should be equal to the natural frequency of the left hand side. (Technically, on a finite time scale you would still see it if the frequencies were just close, but I don't think your source is being "tricky" in this way.) – Ian Mar 7 '17 at 20:11
• @Ian Why is the frequency of right hand side(external force) equal to the left hand side frequency? PS: I have no physics background. – Little Rookie Mar 7 '17 at 20:14
• Mathematically it's because the particular solution to $y''+ay=\cos(bt)$ looks like $c \sin(bt)+d \cos(bt)$ unless $b=a$, and when $b=a$ the particular solution has an additional factor of $t$ which causes the amplitude to grow. – Ian Mar 7 '17 at 20:16

## 1 Answer

3) your possible solutions to a 2nd order ODE with constant coefficients are

$y= C_1 e^{at} + C_2 e^{bt}$

or

$y= C_1 e^{at}\cos bt + C_2 e^{at}\sin bt = C e^{at}\cos (bt + \phi)$

Which of those graphs are possible for each of those.

Question 4.

The solutions for the equations on the left could be

$y = C_1e^{kt} + C_2e^{-kt} + C_3 \cos bt$

or $y = C_1\cos kt + C_2\sin kt + C_3 \cos bt$

or $y = C_1\cos kt + C_2\sin kt + C_3 t \cos kt$

Which one of those might describe the graph.

Wich equation would be associated with that solution.

Update:

$y'' = -y + cos 4t$

Here we have a diff eq where the period of the homogeneous solution is not the same as the period of the particular solution.

$y_g = C_1 \cos t + C_2 \sin t$

$y_p = A \cos 4t + B \sin 4t\\ y'' = -16A \cos 4t - 16B \sin 4t\\ y'' + y = -15A \cos 4t - 15B \sin 4t = \cos 4t\\ y = C_1 \cos t + C_2 \sin t - \frac {4}{15} \cos 4t$

Compare to:

$y'' = -y + cos t$

Where we have a diff eq where the period of the homogeneous solution is the same as the period of the particular solution.

$y_p = A cos t$ can't be the particular solution be because it is already incorporated in the general solution. So, what do we do about that?

$y_p = A t \cos t + B t \sin t\\ y' = A \cos t - A t \sin t + B \sin t + B t \cos t\\ y''= -2A \sin t - A t \cos t + 2B \cos t - B t \sin t\\ y'' + y' = -2A \sin t + 2B \cos t = \cos t\\ y = C_1 \cos t + C_2 \sin t + \frac 12 t \sin t$

To your problem #4)

If the amplitude is growing, the period of the pariticular solution must be the same period as the homogeneous solution.

• For qn 4, im aware of the possible forms of the solution, but knowing that the period of the particular solution may not be equal to the period of the $sin$ and $cos$ function of the general solution for the homogeneous part confuses me. – Little Rookie Mar 7 '17 at 20:23
• The amplitude is growing with time, which means that the form of the equation must be the 3rd type listed.... which means that the period of the particular solution is the same as the period of the homogeneous solution. – Doug M Mar 7 '17 at 20:26
• Why in the 3rd type listed, the period are the same? – Little Rookie Mar 7 '17 at 20:28
• For question 3, B) is right. Why is A) not right? It seems to be the case where the 2 roots are negative. – Little Rookie Mar 7 '17 at 20:37
• I added some more detail regarding question 4, and why the periods must be in synch. That green line doesn't look like the sum of a pair of exponential. It also doesn't look like it could be formed as a linear combination of the other two curves. – Doug M Mar 7 '17 at 20:49