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2nd order ODE - undetermined coefficients?

  1. Apr 22, 2014 #1
    1. The problem statement, all variables and given/known data
    1) Find the general solution of y''+ω02=Ccos3(ωx)

    2) Show there exists two frequencies at which resonance occurs and determine them

    3. The attempt at a solution
    I've tried the method of undetermined coefficients, assuming a solution of the form y=(Acos(ωx)+Bsin(ωx))3. I'm not 100% sure about this - if the RHS was just one trigonometric function, then you assume a solution that's a linear combination of sines and cosines. The trig function here is cubed so I just assumed you cube the linear combination.
    Is this correct the correct approach? Algebra is real hairy if it is!

    For the second part, what exactly is resonance in this context - what am I supposed to show?
     
  2. jcsd
  3. Apr 22, 2014 #2

    AlephZero

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    I guess you made a typo in the equation and it should be ##y''+\omega_0^2 y = C\cos^3(\omega x)##.

    The algebra is probably easier if you write ##\cos^3(\omega x)## as a Fourier series. (The series only has 2 non-zero terms).

    Resonance means the same as in any other context. There are two values of ##\omega## where the general solution has a different form and the amplitude increases to infinity - just like the general solution of ##y''+\omega_0^2 y = C\cos(\omega x)## for one value of ##\omega##.
     
  4. Apr 22, 2014 #3
    Ah yeah, thanks for pointing that out. We haven't done Fourier series yet - I don't think it's even in the syllabus for this course but I'll have a look into it.
     
  5. Apr 22, 2014 #4

    HallsofIvy

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    You don't need to think of it as "Fourier series". Just use trig identities.

    [itex]cos(2x)= cos^2(x)- sin^2(x)= 2cos^2(x)- 1[/itex] so that [itex]cos^2(x)= (1/2)(cos(2x)+ 1)[/itex]. Then [itex]cos^3(x)= (cos^2(x))cos(x)= (1/2)(cos(2x)+ 1)cos(x)= (1/2)cos(x)cos(2x)+ (1/2)cos(x)[/itex].

    And cos(a+b)= cos(a)cos(b)- sin(a)sin(b) so that cos(3x)= cos(2x+ x)= cos(2x)cos(x)- sin(2x)sin(x) while cos(x)= cos(2x- x)= cos(2x)cos(-x)- sin(2x)sin(-x)= cos(2x)cos(x)+ sin(2x)sin(x). Adding those two equations, the "sin(2x)sin(x)" terms cancel giving cos(3x)+ cos(x)= 2cos(2x)cos(x) so that cos(2x)cos(x)= (1/2)cos(3x)+(1/2) cos(x).

    Putting that into the previous equation, [itex]cos^3(\omega x)=(1/2)((1/2)cos(3\omega x)+ (1/2)cos(\omega x))+ (1/2)cos(\omega x)= (1/4)cos(3\omega x)+ (3/4)cos(\omega x)[/itex]. Those are the "two non-zero terms" AlephZero referred to.
     
  6. Apr 22, 2014 #5

    AlephZero

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    The reason I said Fourier series was to give some motivation for what HallsofIvy did. The left hand side of the ODE is the equation for simple harmonic motion, and you might already know about forced vibration and resonance if the right hand side was a single sine or cosine function. So converting the ##\cos^3## into a sum of cosines seems like a good thing to try ....

    Otherwise, it might look a bit like a conjuring trick that you couldn't be expected to figure out for yourself.

    But if you didn't know about what a Fourier series was, communication failure!
     
  7. Apr 22, 2014 #6
    Wow, thanks a lot! Would never have thought of that - trig was one of my extra weaker points back in high school.

    I've solved the non-homogenous case, gotten the resonance frequencies and now doing the homogenous case to get the general solution. The solution I've got for [itex]y''=\omega_0^2y=0[/itex] is [itex]y=iC_1e^{-i\omega_0t}+C_2e^{i\omega_0t}[/itex]. Using Euler's formula, I expanded this out:
    [itex]y=iC_1cos(w_0t)+C_1sin(w_0t)+C_2cos(w_0t)+iC_2sin(w_0t)[/itex].

    There's a theorem that says that if the real and imaginary parts of a solution are also solutions (in this context), so then it would be correct to take the real part

    [itex]y=C_1sin(w_0t)+C_2cos(w_0t)[/itex]

    and add on one of the inhomogeneous solutions to obtain the general solution, right?
     
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