6-01(b)

Problem in compendium:

Find general solution of $LaTeX: \displaystyle \ddot x+2\dot x+5x=0.$

Additional question(s): explain how to solve inhomogeneous eq's.

Three alternative right-hand sides were given.

Recap of the general theory needed:

The general solution of the linear diff.eq. is
C₁u₁+C₂u₂+u* where
- u*=u*(t) is one particular solution, and
- u₁=u₁(t) and u₂=u₂(t) are two non-proportional solutions of the corresponding homogeneous equation (i.e. the one obtained by replacing f by 0).
You are only expected to be able to solve from scratch the cases where a and b are constants, and f is "on a shortlist of doable function forms".
If so, you proceed as follows.
For the constant-coefficients homogeneous equation , replace the second/first/zeroeth derivative by r², r¹, r⁰to get the characteristic equation: . From the formula, the roots are .
The book yields a formula for each of the three cases:
- Two distinct real roots
- One double root (necessarily real!)
- No real roots, but two (distinct) complex roots.
Then for the inhomogeneous equation with nonzero f: try a suitable function form for u* and fit coefficients.

6-01(b) as in the compendium has f=0.

The compendium question

Characteristic equation:
Solution formula yields $LaTeX: r=-1\pm\sqrt{(-1)^2-5\;}=-1\pm\sqrt{-4}$
Non-real roots, but we have a real part of -1, and then that square root can be written as . The "-1" and "2" will enter the solution as follows:
- (Oh, just in case you wondered:
  I said that the square root of 4 is 2. But there is one more: -2i is also an imaginary square root of -4, so why use "2" and "-2"?
  It does not matter. cos(-2t) equals cos(2t), A sin(-2t)=-A sin(2t) is also an arbitrary constant times sin(2t).
  )

OK, so 6-01(b) is solved as in the compendium, and the solution p. 35 obtained. On to the additional questions:

6-01(b), the additional questions from the problem: inhomogeneous equations.

6-01 (b); the equation given is homogeneous and has solution given on page 35.
For the next question ("explain what you would do if the right-hand side were replaced by").
- (i) t² : as particular solution, try a general quadratic and fit coefficients. Note, you cannot hope for just the "k₂t²" to work.
- (ii) t²+3e^t : try and fit coefficients.
- (iii) pick the one you find easiest among e^-2t cos(t) or e^-tcos(2t).
  Here, avoid the one that solves the homogeneous (then you need to try to multiply everything by t). So try the former. Note, you need both cos and sin except special cases.
  This was a bit of talk, you probably want something that would be full score for such an "explain" question on the exam:
  "I choose e^-2t cos t. Try the form $LaTeX: u^*(t)=e^{-2t} (R \cos t+S\sin t)$ and fit R and S, and add u* to the general solution of the homogeneous equation."

So this would be enough to answer such a question. If you need more elaboration, then below the calculations are carried out, and I have explained it in the audio comment (3 minutes). But to be clear: the above is already enough, and the following just elaborates. Like, don't fall into the trap of saying "Kt²" or "Ke^-2tcos t"
For the calculations of (iii) carried out (if you are interested - to repeat, the problem statement did not ask you to do it!):
First, some useful more general calculations. Let u*=vw with $LaTeX: v=e^{\alpha t}$ and $LaTeX: w=R\cos (\beta t)+S\sin(\beta t)$ , and we we have the "easy" derivatives: $LaTeX: \dot v=\alpha v,\quad \ddot v=\alpha^2v,\quad \ddot w=-\beta^2w$ . Also, $LaTeX: \dot w=\beta\cdot(-R\sin(\beta t)+S\cos(\beta t))$ , but let us not insert for that yet, and rather calculate generally as far as we get. We need to add the following terms together:
LaTeX: 5u^*=5vw
$LaTeX: 2\dot u^*=2\big[\dot vw+v\dot w\big]=2v\big[\alpha w+\dot w\big]$
$LaTeX: \ddot u^*=\ddot vw+2\dot v\dot w+v\ddot w=\alpha^2vw+2\alpha v\dot w-\beta^2vw$
The sum is v times the following: $LaTeX: 5w+2\alpha w+2\dot w+\alpha^2w+2\alpha\dot w-\beta^2w=(5+2\alpha+\alpha^2-\beta^2)w+2(1+\alpha)\dot w$
... and since it should be matched to v cos t = e^-2t cos t , put alpha=-2 and beta=1 and insert:
$LaTeX: \cos t=(5+2(-2)+(-2)^2-1^2)w+2(1+(-2))\dot w=4w-2\dot w,$ which when we insert, becomes
$LaTeX: 4R\cos t+4 S \sin t-2\cdot 1\cdot(-R\sin t+S\cos t)=(4R-2S)\cos t+(4S+2R)\sin t$
Matching the sin coefficient to 0 yields 2R=-4S, and matching the cos coefficient 4R-2S=-8S-2S to 1 yields S=-1/10. So if I have done the algebra correctly, I have admittedly not checked thoroughly, it should end up as $LaTeX: u^*=\tfrac1{10}e^{-2t}(2\cos t -\sin t)$