Problem A. Do part (iii) below, about the product of two odd functions. You are not required to do parts (i) and (ii).
(i) (even function)(even function)=even function
(ii) (even function)(odd function)=odd function
(iii) (odd function)(odd function)=even function
Note.These rules are not the same as for multiplying odd and even integers. Rather, they are analogous to the rules for adding odd and even integers. The essential reason is that raising x to an odd power gives an odd function, and raising x to an even power gives an even function, and multiplying the functions means that you add the exponents.

Problem B. Use 9.3 #17 and Problem A parts (a)' and (b)' to show:
(i) if f is an even function then int_-pi^pi f(t) cos(nt) dt = 2 int₀^pi f(t) cos(nt) dt, and int_-pi^pi f(t) sin(nt) dt = 0.
(ii) if f is an odd function then ...[your task is to state some analogous results, but you are not required to prove them].

9.3 #5 - do not compute the series. Instead, your task is to use the sine series given in the answer at the back of the book to find two series for pi (the two series must be different). Write out 9 terms in each series for pi, to make the pattern clear.
Hint. You can use that the coefficients b_n in the sine series are: 2/(n pi) when n=1,7,13,..., and -4/(n pi) when n=3,9,15,..., and 2/(n pi) when n=5,11,17,...

9.4 #2. Then also:
(a) Plot the particular solution x_p(t) and the forcing function F(t) on the same graph. (Either use your graphing calculator, or use Iode with initial conditions x(0)=0.6,x'(0)=0, for forcing function F(t)=3sign(cos(pi t/2)). The initial condition 0.6 has been chosen to approximately equal x_p(0). Make sure you use Runge-Kutta with step size 0.05, to get an accurate solution. Do not use Euler! Use "Plot arbitrary function" to plot the forcing function.)
(b) Estimate the dominant period in the response x_p(t), by using your graph. Compute w_p=2pi/(dominant period).
(c) Compare your frequency w_p with the natural frequency of the system. If they are close, then explain why this has happened, using as evidence the formula for the coefficient a_n in your Fourier series for x_p(t).
[APOLOGY: ignore part (c) of the problem! The dominant period in the graph of x_p(t) is 4, so w_p=pi/2. The natural frequency is sqrt(10), which is not close to pi/2. I was thinking that the term with n=2 would dominate the solution (because putting n=2 into the formula for a_n would give a very small denominator), which students could then explain by noting that the n=2 term in F(t) is close to resonance because its frequency 2pi/2=pi=3.14... is very close to the natural frequency sqrt(10)=3.16... My mistake was to forget that in this particular problem, n is odd and so a_n=0 for all even n, in particular a₂=0. So the dominant term in the response turns out to be the term with n=1, corresponding to frequency pi/2 and period 4.]

9.4 #18. Do not use formula (16). Instead, derive your answer from first principles: find the Fourier series for F(t), and substitute it and a Fourier series for x_p(t) into the differential equation, and then determine the coefficients in x_p(t). (Note your Fourier series for x_p(t) will need both sine and cosine terms.)
(a) For n=1,2,3,4,5,6,7, evaluate the amplitude (a_n²+b_n²)^1/2 of the coefficients. For which n is the amplitude largest?
(b) Compare that dominant frequency n with the practical resonance frequency w_max of the system (see Sec. 3.6 #27). Briefly discuss your findings.

3.5 # 6, 8, 13, 28, 34, 43, 50
and an optional no-credit problem: 59
Hints
For some of these problems, you will want to use the comment at the bottom of the Undetermined Coefficients handout about what to do if f(x) on the right side of the DE consists of a sum of more than one term.
You can check your answers with Wolfram Alpha.
For 43(a), the suggestion in the textbook can be expanded to say: cos(3x) + i sin(3x) = e^i(3x) = (e^ix)³ = (cos(x) + i sin(x))³.
For 50, you should use the Variation of Parameters formula (33) on page 209; you do not have to go through the procedure with u₁ and u₂ on pages 208-209.
For optional problem 59: the DE is equidimensional, but it has equal "roots". That is why the logarithm is needed in y_c.

Feedback: download the Feedback on Iode Project II.

Feedback on 1.4 #66:
(a) [Modeling] Let t=0 be when the snow starts to fall. Consider the short time interval from t to t+dt. During that time, the amount (volume) of snow cleared by the plow equals c dt where the constant c is the rate of snow clearance. On the other hand, the volume of snow cleared by the plow during that time interval can also be expressed as (bt)w dx, where dx is the distance traveled by the plow, w is the width of the plow and bt is the depth of snow on the road (remember the snow falls at a constant rate, so the depth of snow on unplowed parts of the road is proportional to t; here b is our constant of proportionality). Setting equal the two expressions for the amount of snow cleared, we find c dt = btw dx, which can be rewritten as the Differential Equation k dx/dt = 1/t with the constant k = bw/c.
Comment. Notice we used a general principle of modeling: consider a small dt or dx, and describe "what happens" during that short interval. If you can find two different expressions for the same thing, then set the two expression equal.
(b) [Solving] The DE can be written dt/dx=kt, so the solution is t=Ce^kx. Let t=a be the time 7:00am. The problem tells you the following three pieces of information: t(0)=a, t(2)=a+1, t(4)=a+3. This gives three equations for the three unknowns a,C,k. Use the equations to show that a=1, so that 7:00am is one hour after the snow started. Conclude it started snowing at 6:00am.

Feedback on 1.4 #48:
Use the given halflife information to determine the decay constants k₁ and k₂ for U-235 and U-238 respectively. Then by page 37 you have the formulas N₁(t)=C₁ exp(-k₁t) and N₂(t)=C₂ exp(-k₂t) for the amounts of U-235 and U-238. The problem says to assume there was the same amount of each when t=0, so C₁ = C₂. And the problem says that at the present time t=T we have 137.7N₁(T) = N₂(T). (Common error: putting 137.7 on the other side of the equation. Check your formula by re-reading the problem - it says that there is a lot more U-238 than U-235, so N₂(T) must be much bigger than N₁(T).) Solve the last equation for T, to get the age of the universe.