# 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.

• 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

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?

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

• 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