%% Edited F2010 5.6-2 ================================== The characteristic equation is r^2+4=0 with roots 2i,-2i and atom list cos 2t, sin 2t. Then xh=c1 cos 2t + c2 sin 2t. A particular solution xp is found from trial solution x=d1 cos 3t + d2 sin 3t by the method of undetermined coefficients. Ans: d1 = 0, d2 = -1. The initial conditions x(0)=x'(0)=0 are applied to the general solution x = c1 cos 2t + c2 sin 2t - sin 3t in order to evaluate c1 = 0, c2 = 3/2. Then x = (3/2) sin 2t - sin 3t is a sum of two harmonic oscillations of frequencies 2 and 3. 5.6-4 ================================== x'' + 25 x = 90 cos 4t, x(0)=0, x'(0)=90 This is a BEATS problem. For another example, and more details, see 5.6-2, above. Expand the solution as x = x1 + x2, where x1 has frequency 5 and x2 has frequency 4. This identifies x1 as c1 cos 5t + c2 sin 5t, a homogeneous solution x_h, and x2 as d1 cos 4t + d2 sin 4t, a non-homogeneous solution x_p. Superposition implies x = x_h + xp. Find x2 by undetermined coefficients and then determine c1, c2 from the initial conditions x(0)=0, x'(0)=90 [not x1(0)=0, x1'(0)=90, which is a common error]. To repeat, find c1, c2 using x(0)=0, x'(0)=90, from x=c1 cos 5t + c2 sin 5t + d1 cos 4t + d2 sin 4t, where d1, d2 are the two values found from undetermined coefficients. 5.6-8 ================================== x'' + 3x' + 5x = -4 cos 5t The book formulas can be used directly, because (-x)'' + 3(-x)' + 5(-x) = 4 cos 5t, and then F_0 = 4, omega=5, m=1, c=3, k=5 in section 5.6 of E&P. It follows that (1) -x(t) = A cos 5t + B sin 5t with A,B given by formula (19). Then (1) implies (2) x(t) = (-A) cos 5t + (-B) sin 5t. The challenge left is to convert this expression to phase-amplitude form x(t) = C cos(5t - alpha) with C=sqrt((-A)^2+(-B)^2) and tan(alpha)=(-B)/(-A). This is tricky only because you must manually adjust the quadrant for alpha from the point (-A,-B). Read section 5.4, especially between equations (11)-(12), to find out what to do. Beware of the advice in section 5.6, which suggests the calculator answer or the calculator answer plus Pi [see (22) in section 5.6], because that advice applies only to the special problem being studied in section 5.6. 5.6-10 ================================== The characteristic equation is r^2+3r+3=0. Because 10i, -10i are not roots of this equation, then no fixup rule is required and the trial solution for f(x)=8 cos 10t + 6 sin 10t is given by x = d1 cos 10t + d2 sin 10t A particular solution xp is found from this trial solution by the method of undetermined coefficients. Ans: d1 = -956/10309, d2 = - 342/10309. Conversion to phase-amplitude form x(t) = C cos(10t-a) is routinely done from the formulas C = sqrt(d1^2+d2^2), C cos a = d1, C sin a = d2, with answers C = 10sqrt(61)/793, a = Pi + arctan(171/478) == 3.49 radians 5.6-12 ================================== The problem: x'' + 6 x' + 13 x = 10 sin(5t) x(0)=0, x'(0)=0 LAPLACE THEORY SOLUTION (Recommended). The equation for L(x) is 50 L(x) = -------------------- (s^2+25)((s+3)^2+4) After a lot of partial fraction details, the answer is found to be A cos(5t) + B sin(5t) + C exp(-3t) cos(2t) + D exp(-3t) sin(2t) A = -25/87, B = -10/87, C = 50/174, D = 125/174 The TRANSIENT solution consists of the two terms with negative exponentials. The STEADY-STATE solution is A cos(5t) + B sin(5t). This is the same as the undetermined coefficients solution documented in the textbook in section 5.6. You cannot use the formulas in the textbook to find A,B, because the term on the right of the differential equation is sin(2t), NOT cos(2t). UNDETERMINED COEFFICIENTS SOLUTION. Substitute x = A cos(2t) + B sin(2t) into the differential equation to get the following identity, which is a competition between two different linear combinations of the independent atoms cos(2t) and sin(2t). (-12 A + 30 B)cos(5t) + (-30 A -12 B)sin(5t) = (0)cos(5t) + (10)sin(5t) Then corresponding coefficients must match, by independence, which gives the 2x2 system for A,B: -12 A + 30 B = 0, -30 A - 12 B = 10. Solve this system by Cramer's rule. The answers are A = -25/87, B = -10/87. PHASE-AMPLITUDE SOLUTION. This conversion uses formulas found in 5.4 of the Edwards-Penney textbook. Then x(t) = F cos(5t - alpha) F = sqrt(A^2+B^2) = 5 sqrt(29)/87 alpha = arctan(B/A) + Pi or 2 Pi Choose alpha=arctan(B/A) + Pi, because (A,B) is in quadrant 3. PLOTS. A replica of the plot obtained, hand-drawn, can be submitted instead of computer printout. A hand calculator is enough of an instrument to produce the plot, using the equation x(t) = A cos(5t)+B sin(5t). The second plot, which adds in the exponential terms, x(t) = A cos(5t)+B sin(5t)+C exp(-3t)cos(2t)+D exp(-3t)sin(2t) can be done the same way. If you did the problem by Laplace, then C and D are already known. If not, then compute them from the preceding 4-term equation for x(t) and the initial conditions x(0)=0, x'(0)=0. 5.6-16 ================================== The equation is x'' + 4x' + 5x = 10 cos(wt). The theory says that practical resonance occurs exactly for input frequency w determined by the equation [E&P problem 5.6-27] w = sqrt((k/m) - (1/2)c^2/m^2) Because m=1, c=4, k=5, then w = sqrt(5-9) This value is NOT REAL, therefore practical resonance does not occur. Contrast this result with 5.6-18, in which w=sqrt(600) and practical resonance does occur. 5.6-18 ================================== The equation is x'' + 10x' + 650x = 100 cos(wt). The theory says that practical resonance occurs exactly for input frequency w determined by the equation [E&P problem 5.6-27] w = sqrt((k/m) - (1/2)c^2/m^2) Because m=1, c=10, k=650, then w = sqrt(600) This value is REAL, therefore practical resonance occurs. ======== end of problem notes 5.6 ========