DE 2

DE 2

1. General Homogeneous Equations

   • Focusing on solving equations of the form:
     • • We require two initial values, and have an underlying assumption of linearity

2. Homogeneous Linear Second-Order ODEs with constant coeffients

   • We expect a solution of the form x(t)=Ax1(t)+Bx2(t)
   • We can generate an auxilliary equation ak^2+bk+c=0,and the solution depends on the roots of this equation
     • • if k<0 the x(t)→0 as t→∞

3. Mass and Spring Systems

   • From mv'=F(t) ie f=ma and f=kx, consider:
     • • Then, with no friction we have mx''=-kx
         • The simplest model for friction is that it is proportional to velocity of the object.
           • hence we have mx''=-kx-cx' ie mx''+cx'+kx=0
             • with c=0 we have simple harmonic motion, giving solutions
               • • • Assing friction give an equation with a solution dependent on the discriminant
                     • Undamped c=0
                       • Without friction the mass oscillates back and forth forever
                     • Underdamped c^2-4mk<0
                       • In this case, the system still oscillates, but the amplitude decreases gradually
                     • Critically damped c^2-4mk=0
                       • In this case roots are -c/2m
                       • We have solution x(t)=(A+Bt)e^-((c/2m)t)
                       • If the mass is released from rest then its distance from equilibrium decreases monotonically to 0, if sufficient momentum is initially imparted then the mass will overshoot before increasing to zero
                     • Overdamped c^2-4mk>0
                       • • This gives similar results to critcal damping, but the decay is less severe

4. Inhomogeneous Linear Second-Order ODEs

   • Consider a(t)x''+b(t)x'+c(t)x=f(t)
     • f(t) is sometimes known as the forcing term
   • Because of linearity, if xp(t) is a solution to the homogeneous case of these equations, it is also a solution to the inhomogeneous equations. This homogeneous case is called the complimentary function and the solution is the particular integral
   • Constant Coefficients
   • Mass and Spring Systems with Forcing
     • • If we now force this oscillation system, then the solutions become much more interesting. If we can push the mass every second, say it may reduce the oscillations, or if we hit a certain frequency, the oscillations may become much larger very quickly. We can approximate the force as Fcos(Ωt), giving us our general equation:
       • Representing a mass/spring system being foced periodically
       • As t→∞, the second term "vanishes". This second term is called the "transient behaviour", while the first term is known as the "steady state solution"
         • With no forcing (F=0) and no friction (c=0) we have the system oscillates with frequency ω/2π the "natural frequency"
           • If there is little or no friction and F=/=0 then we look at what happens as we change Ω- in particular, what happens as Ω gets close to Ω→ω. As ω the amplitude A gets larger and larger. In fact, if c=0, then as Ω→ω, A→∞ . This is known as resonance.
             • Our solution is of course wrong if we take Ω=... and c=0. In this case the ansatz for the particular integral (Acos(Ωt)+Bsin( Ωt)) can not be used because it solves the homogeneous case. So we have to try something else. It turns out that xp(t)=Atcos( Ωt)+Btsin( Ωt) works and this shows that in the resonance case we have an amplification of the amplitude of the oscillation which grows uncontrolled as t→∞. Of course what really happens is that our model assumptions are no longer valid

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