Edwards-Penney Chapter 3, sections 3.5, 3.6, 3.7 Topics, Definitions, Examples, Theorems 3.5 Non-homogeneous differential equations and undetermined coefficients ==== PROJECT: Find yp in the superposition formula y=yh+yp for the equation an D^ny + ... + a1 Dy + a0 y = f(x). METHOD of UNDETERMINED COEFFICIENTS The method applies to nth order equations with constant coefficients. The RHS=f(x) is required to be A COMBINATION OF SOLUTION ATOMS. EXAMPLE. Find yp for y'' + 3y' + 4y = 3x + 2 ANSWER: yp=Ax+B, A=3/4, B=-1/16 METHOD: Substitute y=Ax+B into the DE and match coefficients of x. This gives 2 equations in the 2 unknowns A,B, which when solved produces the unique solution reported. EXAMPLE. Find yp for y'' - 4y = 2 exp(3x) ANSWER: yp=A exp(3x), A=2/5. METHOD: Substitute y=A exp(3x) into the De, get one equation for the unknown A, by matching coefficients of exp(3x). The unique solution is A=2/5. EXAMPLE. Find yp for y'' - 4y = 2 exp(2x) ANSWER: yp=A x exp(2x), A=1/2 METHOD: Argue that yp=A exp(2x) won't work because substitution of y=A exp(2x) produces zero on the left which cannot match 2 exp(2x) on the right. Without much motivation the TRIAL SOLUTION is guessed to be y=A x exp(2x). Substitution and matching coefficients of like terms gives one equation in one unknown A, with unique solution A=1/2. RULE 1 for the METHOD of UNDETERMINED COEFFICIENTS Suppose that no term from f, f', f'', etc, satisfies the associated homogeneous differential equation. Then a trial solution is a linear combination of all independent terms extracted from f, f', f'' (all derivatives of f(x)). The coefficients are found by substitution of the trial solution into the non-homogeneous equation. EXAMPLE. Find yp for y'' + 4y = 3 x^3 ANSWER: yp=Ax^3 + Bx^2 + Cx + D, A=3/4, B=0, C=-9/8, D=0. EXAMPLE. Find yh and yp for y'' - 3y' + 2y = 3 exp(-x) - 10 cos(3x), then find y for initial conditions y(0)=1, y'(0)=2. ANSWER: yh=c1 exp(x) + c2 exp(2x) from characteristic equation r^2-3r+2=0. yp=A exp(-x) + B cos(3x) + C sin(3x) because f(x) and all its derivatives are combinations of such terms. Substitution and coefficient-matching on the terms gives A=1/2, B=7/13, C=9/13. Finally, y=yh+yp has two coefficients c1, c2. They are determined from the two initial conditions y(0)=1, y'(0)=2 by solving a system of 2 equations in 2 unknowns: c1=-1/2, c2=6/13. EXAMPLE. Find yh and the form of a particular solution yp for y''' + 9y' = x sin(x) + x^2 exp(2x) ANSWER: Characteristic equation r^3+9r=0 gives yh = c1 + c2 cos(3x) + c3 sin(3x). The trial solution is a linear combination of the terms cos x, x cos x, sin x, x sin x, exp/(2x), x exp(2x), x^2 exp(2x). DUPLICATION and RULE 1 The case when some terms in f(x) and its derivatives appear in the homogeneous solution yh causes RULE 1 to fail. But not completely: RULE 1 predicts the correct number of terms, which is the number of constants to be determined. Incorrect terms from RULE 1 are off by a power of x. Exponential and trig factors in a term predicted by RULE 1 are correct. How to fix the RULE 1 trial solution, when the rule fails? Repair the power of x for the affected terms only. Details below. EXAMPLE. Find the form of yp for (D-5)^3 y = (2x-3) exp(5x) ANSWER: A linear combination of x^3 exp(5x), x^4 exp(5x). RULE 1 predicts terms exp(5x), x exp(5x) for the trial solution. Already known is that there will be 2 terms and 2 coefficients A,B to be found in the method of undetermined coefficients. THE CATCH: RULE 1 does not apply, because both predicted terms are solutions of the homogeneous equation. The textbook EXPLANATION uses root multiplicities to replace the RULE 1 terms with ones that work. The replacement terms are the RULE 1 terms multiplied by a certain power of x. RULE 2 for the METHOD of UNDETERMINED COEFFICIENTS Find the terms for the trial solution from RULE 1. This is the correct number of terms. A term will be unmodified if it is not a solution of the homogeneous equation. A term that is a solution of the homogeneous equation will be modified by multiplication by a power of x. The method is best learned by example, using Euler's theorem for solution atoms in reverse to predict the power. THE BASIC RULE: Terms corresponding to the same root in Euler's theorem get multiplied by the same power of x, that power being the least one needed to keep all terms being multiplied from being a solution of the homogeneous equation. Most readers find Figure 3.5.1, a table of choices for yp, too mysterious for daily use because of the mystery x^s appearing in the formulas. EXAMPLE. Find yp for y''' + y'' = 3 exp(x) + 4 x^2 ANSWER: A linear combination of exp(x), x^2,x^3,x^4. Then yp=A exp(x)+Bx^2+Cx^3+Dx^4, A=3/2, B=4, C=-4/3, D=1/3. EXPLANATION. The trial solution by RULE 1 is a linear combination of the terms exp(x), 1, x, x^2. All are solutions of the homogeneous equation so RULE 1 fails for all terms. Replacement occurs in groups as follows: Group 1: exp(x) corresponding to Euler root=1 Group 2: 1,x,x^2 corresponding to Euler root=0 The replacements for each group are the same terms multiplied by a power of x. Group 1 is unmodified (power x^0). Group 1: exp(x) Group 2 replace: x^2,x^3,x^4 [power x^2] The reasoning in each group is to multiply every term in the group by the same power of x, choosing the least power that will make all group terms not a solution of the homogeneous equation y''' + y'' = 0. Groups not violating RULE 1 are unmodified, or if you wish, they are multiplied by x^0. EXAMPLE. Find the form of yp for y''+6y'+13y=exp(-3x)cos(2x) ANSWER: yp is a linear combination of terms x exp(-3x)cos(2x), x exp(-3x)sin(3x). EXAMPLE. Find the form of yp for (D-2)^3(D^2+9)y=x^2 exp(2x)+x sin(3x) ANSWER: yp is a linear combination of the terms x^3 exp(2x), x^4 exp(2x), x^5 exp(2x) x cos(3x), x^2 cos(3x), x sin(3x), x^2 sin(3x). EXPLANATION: RULE 1 predicts terms Group 1: exp(2x), x exp(2x), x^2 exp(2x) Group 2: cos(3x), x cos(3x), sin(3x), x sin(3x) The trial solution will have 7 terms obtained by modifying groups 1 and 2 with powers of x. Multiply group 1 by x^3 and group 2 by x. They are the least powers which render the replacement terms not a solution of the homogeneous equation. VARIATION of PARAMETERS The theory find yp(x) for A(x)y''+B(x)y'+C(x)y=f(x) when A,B,C,f are continuous and A is never zero. THEOREM 1. Variation of Parameters The equation y'' + P(x)y' + Q(x)y = f(x) with homogeneous solution yh=c1 y1(x) + c2 y2(x) has particular solution yp(x)=D1 y1(x) + D2 y2(x) where D1=-int(y2 f/W,x), D2=int(y1 f/W,x), and W = y1 y2' - y1' y2 is the Wronskian of y1, y2. REMARK. If starting with Ay''+By'+Cy=f, and A is not 1, then replace W by AW in the formulas, to adjust for division by A. EXAMPLE. Solve for yp(x) in y''+y=tan(x) ANSWER: yp(x)= -cos(x) ln|sec(x) + tan(x)| EXPLANATION. The variation of parameters formula suggests to find the general solution of y''+y=0 (homogeneous equation) and then identify independent solutions y1, y2 from the answer. We get y1=cos(x), y2=sin(x) and Wronskian W=1. Then D1=-int(y2 f/W,x)=-int(sin(x)tan(x),x) =sin(x)-ln|sec(x)+tan(x)| [integral table] D2=int(y1 f/W,x)=int(cos(x)tan(x),x) =-cos(x) The constants of integration were dropped, because it would only add a linear combination of y1, y2. Then yp= D1 y1 + D2 y2 = (sin(x)-ln|sec(x)+tan(x)|)cos(x)+(-cos(x))sin(x) = - cos(x) ln|sec(x)+tan(x)| REMARK on the EXERCISES The 3.5 exercises on variation of parameters sometimes give y1, y2 instead of asking you to solve the homogeneous equation. They want equations with non-constant coefficients, but you couldn't know how to solve them from sections before 3.5. So they solve them for you. 3.6 Forced oscillations and resonance ==== SPRING-MASS FORCED OSCILLATOR m x'' + c x' + k x = F(t) UNDAMPED FORCED OSCILLATOR Front-loading washing machine example. Rotating cart, undamped: m x'' + k x = F0 cos(omega t), F0=m0 a omega^2 (a constant) DEF. omega0=sqrt(k/m) is the NATURAL FREQUENCY of the oscillator. DEF. The FREQUENCY of a trig oscillation with natural frequency omega is 2 Pi/omega. PARTICULAR SOLUTION when omega unequal to omega0 xp(t) = (F0/m)cos(omega t)/(omega0^2 - omega^2) Found by undetermined coefficients EXAMPLE. Solve x'' + 9 x = 80 cos(5t), x(0)=x'(0)=0. ANSWER: x(t)=5 cos(3t) - 5 cos(5t) DETAILS: xh=c1 cos(3t) + c2 sin(3t) has frequency 3 unequal to the frequency omega=5 of the input term 80 cos(5t). Then xp(t) = (F0/m)cos(omega t)/(omega0^2 - omega^2) = 80 cos(5t)/(9-25)=-5 cos(5t) x(t) = xh + xp = c1 cos(3t)+c2 sin(3t)-5 cos(5t). Substitute x(t) into conditions x(0)=0, x'(0)=0 to obtain c1=5, c2=0. BEATS THEOREM. Solve m x'' + k x = F0 cos(omega t), x(0)=x'(0)=0, then x(t)= C(cos(omega t) - cos(omega0 t)), C=(F0/m)/(omega0^2 - omega^2) THEOREM. cos(2a)-cos(2b) = 2 sin(a-b) sin(a+b) implies cos(omega0 t) - cos(omega t) = 2 sin(At) sin(Bt) where A=(omega0-omega)/2, B=(omega0+omega)/2 DEF. BEATS is a solution proportional to cos(omega0 t) - cos(omega t) where the two frequencies are different. HISTORY. It refers to a physical phenomenon described in physics texts. For example, two musical instruments played with nearly matching frequencies will produce an audible variation in the amplitude of the sound, a BEAT. EXAMPLE. An oscillation created by superposition of two different frequencies can show BEATS: x(t)=cos(55t)-cos(45t)=2sin(5t)sin(50t) with SLOWLY VARYING amplitude function 2sin(5t) and RAPIDLY VARYING oscillation sin(50t). RESONANCE THEOREM. If omega0=omega, then all solutions of mx''+kx=F0 cos(omega t) are unbounded. DEF. The response x(t) of an undamped forced system mx''+kx=F0 cos(omega t) to frequency-matching omega0=sqrt(k/m)=omega results in all solutions being unbounded as t goes to infinity. This is called PURE RESONANCE. We summarize this phenomenon by saying that the external force ADDS to the AMPLITUDES of NATURAL VIBRATIONS. EXAMPLE. Rotating cart example. Assume 5x'' + 500x = F0 cos(omega t) models the rotating cart. Frequency matching is sqrt(500/5)=omega in radians per second. All solutions contain the undetermined coefficients solution xp = (F0/10) t sin(10 t) [assumed omega=10] which is unbounded at t goes to infinity. The other terms in the solution are bounded harmonic functions of frequency 10. MODELING MECHANICAL SYSTEMS Kinetic energy T = (1/2) m v^2 a mass m moving with velocity v. Potential energy V = (1/2) k x^2 for a spring with Hooke's constant k undergoing elongation x. Gravitational potential energy V = m g h for a mass m at height h above the reference level V=0, g=gravitational constant. Rotational kinetic energy I = (1/2) I omega^2 for a rotating body at angular velocity omega and moment of inertia I. EXAMPLE. Mass m is a uniform disk of radius a attached to a spring of Hooke's constant k, rolling without slipping. Find its natural frequency omega0. ANSWER: omega0=sqrt((2/3)(k/m)). It is equivalent to mass m oscillating undamped on a spring with Hooke's constant k reduced by 2/3. METHOD: Add kinetic mv^2/2 energy and rotational kinetic energy (1/2) I omega^2 and the spring potential energy (1/2) k x^2 to get a constant E, the total energy. Insert I=ma^2/2 for a uniform disk and v=a omega. Then (3/4)m v^2 + (1/2)k x^2 = E. Differentiate on t to obtain DE x'' + b x = 0 with b=(2/3)(k/m). Then omega0=sqrt(b). EXAMPLE. Car on a washboard road. Find resonant speed v. MODEL: mx'' + kx = ka cos(2 Pi v t/L) m=800, k=70000, a=5/100, L=10, v=unknown car speed ANSWER: v = 14.89 m/s = 33.3 mi/h METHOD: Match frequencies, sqrt(k/m)=2 Pi v/L. All symbols in this equation are known, except for v. DAMPED FORCED OSCILLATIONS Consider mx'' + cx' + kx = F0 cos(omega t) THEOREM. The solution xh(t) of the homogeneous problem mx'' + cx' + kx = 0 satisfies for m>0, c>0, k>0 limit xh(t) = 0 at t=infinity. DEF. A TRANSIENT SOLUTION is one that dies out with passage of time, that is, its limit at infinity is zero. DEF. A STEADY-STATE solution (or OBSERVABLE solution) is an expression in which all terms limiting to zero at infinity have been removed. We use the term when it is a solution of the DE. THEOREM. The equation mx'' + cx' + kx = F0 cos(omega t) has undetermined coefficient solution xp(t) = A cos(omega t) + B sin(omega t) where A,B are given by A = (k-m omega^2)F0/delta, B = c omega F0/delta, delta = (k-m omega^2)^2+(c omega)^2 DEF. The AMPLITUDE C(omega) of the steady-state solution xp(t) (see above) is defined by F0 C(omega) = sqrt(A^2+B^2)= --------------------------------- sqrt((k-m omega^2)^2+(c omega)^2) THEOREM. The phase-amplitude form of the steady-state solution is given by xp(t) = C(omega) cos(omega t - alpha) where tan(alpha) = B/A = c omega/(k - m omega^2) EXAMPLE. Consider the problem x'' + 2 x' + 26 x = 82 cos(4t), x(0)=6, x'(0)=0 Find the transient and steady-state solutions and report on practical resonance. ANSWER: xss(t)=xp(t)=sqrt(41) cos(4t-alpha), tan(alpha)=4/5. xtr(t)=exp(-t)(cos(5t)-3 sin(5t)) Practical resonance occurs for omega=sqrt(24). DETAILS: The steady-state is xp(t) from the Theorem. The transient is the collection of negative exponential terms appearing in the solution, part of the homogeneous solution. THEOREM. Practical resonance occurs for omega = sqrt(k/m - (1/2)c^2/m^2) Ref: See exercise 27. We make sense of this formula only when the argument of the square root is positive; otherwise, we say only that practical resonance does not occur. 3.7 Electric Circuits ==== RLC Circuit diagram Derivation L (DI/dt) + R I + Q/C = E(t) Electrical-Mechanical Analogy DC voltage AC current Transient current Steady-state periodic current I(t) = A cos(omega t - alpha) Phase angle alpha = arctan( omega R C /(1 - L C omega^2) ) Amplitude A = E0/Z Z^2 = R^2 + S^2 S = omega L - 1/(Omega C) is called the REACTANCE Z is called the IMPEDANCE Time Lag delta = arctan(S/R) time lag = delta/omega seconds The connection between input voltage E(t) and output steady-state current I(t) is the following. I(t) = E0/Z * cos(omega t - alpha) = E0/Z * sin(omega t - delta) E(t) = E0 * sin(omega t) = input voltage EXAMPLE 1. Find the steady-state current and the time lag for the model 0.1 * I'' + 50 * I' + 2000 * I = I(0)=0, Q(0)=0 The value omega = 377 is approximately 2*Pi*60, for 60 Hz alternating current. Value 110 is for a 110 volt home supply. The TRICK: Use 0.1 I'(0) + R I(0) + Q(0)/C = E(0) to determine I'(0). Then solve the equation for I(t) using standard methods. EXAMPLE 2. Find the steady-state current and the time lag for the model 0.1 * I'' + 50 * I' + 2000 * I = 0 I(0)=0, Q(0)=0 which is the result of a DC input voltage E(t)=110. The TRICK: Use 0.1 I'(0) + R I(0) + Q(0)/C = E(0) to determine I'(0). Then solve the equation for I(t) using standard methods. In this example, I'(0)=E(0)/0.1=1100. ELECTRICAL RESONANCE The notion of resonance is that the amplitude of the oscillation is as large as possible. The amplitude is E0/Z, which a maximum when Z is a minimum. This happens exactly when S^2=0, which makes Z^2 = R^2 + S^2 reduce to R^2, or Z=R. The largest amplitude is E0/R. CRITICAL FREQUENCY This is the value of omega which makes the amplitude a maximum E0/R. We compute omega = 1 / sqrt(LC). CRYSTAL RADIO TUNER AM RADIO TUNER FM RADIO TUNER ========end