Chapter 3 Problem Notes Section 3.1 ===================== Updated 2015 3.1: 34, 36, 38, 40, 42, 46, 48 3.1-34 ====== y'' + 2y' - 15y = 0 r^2 + 2r -15=0, char equation (r+1)^2 -16=0, complete the square (r+1-4)(r+1+4)=0, factor difference of squares r=3, r= -5, find the two roots [could have used the quadratic formula] exp(3x), exp(-5x), the two atoms implies by Euler's theorems y = linear combination of the atoms y = c1 exp(3x) + c2 exp(-5x) 3.1-36 ====== 2y'' + 3y' = 0 2r^2+3r=0, char equation r(2r+3)=0, factored r=0, r=-3/2, roots exp(0x), exp(-3x/2), two atoms implied by Euler's theorems y=linear combination of the atoms y=linear combination of 1, exp(-3x/2) y = c1 + c2 exp(-3x/2) 3.1-38 ====== 4y'' + 8y' + 3y = 0 4r^2+8r+3=0, char equation (2r+1)(2r+3)=0, factored by inverse FOIL r=-1/2, r=-3/2, roots exp(-x/2), exp(-3x/2), atoms by Euler's theorems y=linear combination of the atoms 3.1-40 ====== 9y'' - 12y' + 4y = 0 9r^2-12r+4=0, char equation (3r-2)(3r-2)=0, factored by inverse FOIL r=2/3, r=2/3, double root) exp(2x/3), x exp(2x/3), two atoms by Euler's theorems y=linear combination of the atoms 3.1-42 ====== 35y'' - y' - 12y = 0 35r^2-r-12=0, char equation (7r+4)(5r-3)=0, factored by inverse FOIL r=-4/7, r=3/5, roots exp(-4x/7), exp(3x/5), two atoms by Euler's theorems y=linear combination of the atoms 3.1-46 ====== y = c1 exp(10x) + c2 exp(100x), find the DE Take partials on c1, c2 across the general solution to find the basis functions: exp(10x), exp(100x) These atoms correspond to Euler roots r=10, r=100 Then (r-10)(r-100)=0 must replicate the characteristic equation. Expand, r^2-110r+1000=0, the charactistic equation Then y'' -110y' + 1000y=0 is the differential equation 3.1-48 ====== y = exp(x)(c1 exp(sqrt(2)x) + c2 exp(-sqrt(2)x)), find the DE y = c1 exp(r1 x) + c2 exp(r2 x) where exp(r1 x)=exp(x)exp(sqrt(2)x)=exp([1+sqrt(2)]x), exp(r2 x)=exp(x)exp(-sqrt(2)x)=exp([1-sqrt(2)]x), imply r1=1+sqrt(2), r2=1-sqrt(2). Like the preceding problem, use the factor-root theorem of college algebra to reconstruct the characteristic equation (r - r1)(r - r2)=0 r^2 - (r1+r2)r + r1 r2=0 r^2 - 2r + 5 = 0 Then the differential equation must be y'' - 2y' + 5y = 0 Section 3.2 ===================== Updated 2015 3.2: 18, 22 3.2-18 ====== y''' - 3y'' + 4y' -2y = 0 is not actually used in this problem. You are not supposed to solve for y, because given are three atoms which generate the general solution y. y=linear combination of three atoms y=linear combination of exp(x), exp(x) cos x, exp(x) sin x y = c1 exp(x) + c2 exp(x) cos x + c3 exp(x) sin x Given initial conditions y(0)=1, y'(0)=0, y''(0)=0, then c1, c2, c3 are uniquely determined. Your problem is to find c1, c2, c3, using linear algebra methods. First, write three equations, one for each of y, y', y'', which involve the symbols c1, c2, c3, x. Substitute x=0 in each equation, then collect on c1, c2, c3 to obtain a system of 3 equations in 3 unknowns. Oberve that substitution x=0 eliminates symbol x from the equations, leaving only three symbols c1, c2, c3. c1 exp(0) + c2 exp(0) cos 0 + c3 exp(0) sin 0 = y(0) = 1, c1 exp(0) + c2 exp(0) (cos 0 - sin 0) + c3 exp(0) (sin 0 + cos 0) = 0, c1 exp(0) + remaining terms similar to above = 0. The system is then of the form Au=b where A:=Matrix([[1,1,0],[1,1,1],[1,*,*]]); b:=<1,0,0>; u:=; Solve it for c1, c2, c3 (the answer is unique). 3.2-22 ====== The equation y''-4y=12 is never used, because both a particular solution y_p = -3 and the general solution y_h of the homogeneous problem y'' -4y=0 has been supplied. Superpositon says that y=y_h+y_p, or y = c1 exp(2x) + c2 exp(-2x) + (-3) The initial conditions y(0)=0, y'(0)=10 apply to the equation immediately above, not to the equation for y_h, which is the common error. As in 3.2-18, differentiate the general solution y, to get two equations, one for y and one for y', each in terms of c1, c2, x. Substitute x=0 on both to eliminate the symbol x, leaving only symbols c1, c2, to obtain c1 exp(0) + c2 exp(0) - 3 = 0, 2 c1 exp(0) - 2 c2 exp(0) + 0 = 10. This is a system Au=b where A:=Matrix([[1,1],[2,-2]]); b:=<3,10>; u:=; Solve it for c1, c2. the answer is unique. Section 3.3 ===================== Updated 2015 3.3: 8, 10, 16, 32, 40 3.3-8 ===== y'' - 6y' + 13y = 0 r^2 - 6r + 13 = 0, char equation (r-3)^3 + 4 = 0, complete the square (r-3 + 2i)(r-3 - 2i) = 0, factor sum of squares with complex numbers roots 3 + 2i, 3 - 2i exp(3x)cos(2x), exp(3x)sin(2x), atoms by Euler's theorems y=linear combination of the atoms 3.3-10 ====== 5 D^4 y + 3 D^3 y = 0 5r^4 + 3r^3 = 0, char equation r^3(5r+3)=0, factored r=0,0,0 and r=-3/5, roots according to multiplicity exp(0x), x exp(0x), x^2 exp(0x) three atoms for the 3 roots r=0,0,0 exp(-3x/5), atom for the root r=-3/5, by Euler's theorems y = linear combination of the atoms 3.3-16 ====== D^4 y + 18 D^2 y + 81y = 0 r^4 + 18r^2 + 81 = 0, char equation u^2 + 18u + 81 = 0, where u=r^2 (u+9)(u+9)=0, factored by inverse FOIL (r^2+9)^2 = 0, substitute u=r^2 r = 3i, 3i, -3i, -3i, roots according to multiplicity cos(3x), x cos(3x), sin(3x), x sin(3x), the four atoms, by euler's theorems y=linear combination of the atoms 3.3-32 ====== D^4 y + D^3 y - 3 D^2 y -5 D y - 2y = 0 r^4 + r^3 - 3r^2 - 5r - 2 = 0, char equation You are supposed to find the roots using Theorey of Equations from college algebra. This involves the following theorems: Root-factor theorem Rational root theorem Division algorithm for polynomials PROCEDURE. Start by applying the rational root theorem to distill the possible rational roots into the finite list of all factors of the constant term (-2) divided by all possible factors of the leading coefficient (1). This gives the root list 2/1, -2/1, 1/1, -1/1 Try each of the four numbers in the characteristic equation, to determine if it is a root. For example, r=1 is not a root, because r^4 + r^3 - 3r^2 - 5r - 2 = 1+1-3-5-2=-8, not zero On the other hand, r=-1 is a root, because r^4 + r^3 - 3r^2 - 5r - 2 = 1-1-3+5-2=0. Then (r-(-1)) is a factor of the characteristic polynomial, meaning r+1 perfectly divided it. Use long division of polynomials to find the quotient after division, which is a cubic. Then r^4 + r^3 - 3r^2 - 5r - 2 = (r+1)(cubic) The project is then reduced to factoring the cubic, which can be attacked by the same set of theory of equations tools cited above. Finally, the roots are known, which happen to be r = -1, -1, -1, 2. Then the atoms are exp(-x), x exp(-x), x^2 exp(-x) and exp(2x) by Euler's theorems. The solution y is a linear combination of the atoms. 3.3-40 ====== y =A exp(2x) + B cos(2x) + C sin(2x), find the DE Like problems in section 3.1, the idea is to differentiate on the symbols A, B, C to identify the basis functions, which are atoms. Then exp(2x), cos(2x), sin(2x) are the three atoms. Euler's theorem produces these atoms from the three roots r=2, r = 2i, r = -2i. Then r-2, r-2i, r+2i are factors of the characteristic equation, by the root-factor theorem of college algebra. Finally, the product of these three factors must give the characteristic equation. From it, we construct the DE. (r-2)(r-2i)(r+2i)=0, (r-2)(r^2+4)=0, r^3 - 2r^2 + 4r - 8 = 0, y''' - 2y'' + 4y' - 8y = 0. Section 3.4 ===================== Updated 2015 3.4-18 (a) 2x'' + 12x' + 50x=0, x(0)=0, x'(0)=-8; solve the equation. ====== Classify as overdamped, critically damped, underdamped. Write x(t) in phase=amplitude form. (b) 2u'' + 0u' + 50u=0, u(0)=0, u'(0)=-8; solve the equation. Write u(t) in phase=amplitude form. (a) 2x'' + 12x' + 50x=0, x(0)=0, x'(0)=-8; The characteristic equation for the damped system is 2(r^2+6r+25)=0 with complex conjugate roots r=-3+4i,r=-3-4i. Solution Atoms=e^{-3t}cos 4t, e^{-3t}sin 4t. Underdamped. (b) 2u'' + 0u' + 50u=0, u(0)=0, u'(0)=-8; The characteristic equation for the undamped system is 2(r^2+0+25)=0 with complex conjugate roots r=5i,r=-5i. Solution Atoms=cos 5t, sin 5t. Solve each homogeneous system, as a linear combination of the solution atoms. Evaluate the constants in the linear combination, in each of the two cases, using the initial conditions x(0)=0, x'(0)=-8. There are two linear algebra problems to solve. Write each solution in phase-amplitude form, a trig problem. Plot the solutions so obtained, on a hand calculator or maple, and draw by pencil a graphic replica next to your hand-written solution. 3.4-20 (a) 2x'' + 16x' + 40x=0, x(0)=5, x'(0)=4; solve the equation. ====== Classify as overdamped, critically damped, underdamped. Write x(t) in phase=amplitude form. (b) 2u'' + 0u' + 40u=0, u(0)=5, u'(0)=4; solve the equation. Write u(t) in phase=amplitude form. 2x'' + 16x' + 40x=0 The characteristic equation for the damped system is 2(r^2+8r+20)=0 with complex conjugate roots r=-4+2i,r=-4-2i. Solution Atoms=e^{-4t}cos 2t, e^{-4t}sin 2t. Underdamped. 2x'' + 0x' + 40x=0 The characteristic equation for the undamped system is 2(r^2+0+20)=0 with complex conjugate roots r=sqrt(20)i,r=-sqrt(20)i. Solution Atoms=cos( sqrt(20)t), sin( sqrt(20)t). Solve each homogeneous system, as a linear combination of the solution atoms. Evaluate constants c1,c2 in the linear combination, in each of the two cases, using the initial conditions x(0)=5, x'(0)=4. There are two linear algebra problems to solve. Answers: Coefficients c1=5, c2=12 for 2x'' + 16x' + 40x=0 Amplitude sqrt(5^2 + 12^2) = 13 Coefficients c1=5, c2=2/sqrt(5) for 2x'' + 0x' + 40x=0 Amplitude sqrt(5^2 + 4/5) = sqrt(129/5) Plot the solutions so obtained, on a hand calculator or maple, and draw by pencil a graphic replica next to your hand-written solution. Don't plot from the phase-amplitude conversions, because with computing assist there is no reason to do that. Write each solution in phase-amplitude form, a trig problem. See section 3.4 for specific instructions on how to find the amplitude C and phase shift alpha. The book's answers are correct. (a) tan(alpha) = 12/5 (b) tan(alpha) = 5 sqrt(5)/2 3.4-34 100 x'' + cx' + kx = 0. Given a solution x(t) such that its ====== first two maxima are t=0.34 seconds, x=6.73 inches and t=1.17 seconds, x=1.46 inches. Find c and k. Use the results freely from problems 32 and 33 to solve problem 34. Let x1=6.73/12 and x2=1.46/12 in problem 33. Use times t1=0.34 and t2=1.17 in problem 32. Problems 3.4-32 and 3.4-33 are applied to this problem to determine the values for symbols p and omega_1; these symbols will now be defined, using the characteristic equation and the quadratic formula. The characteristic equation: m r^2 + c r + k = 0 Because of the oscillation described in 3.4-34, this has to be the underdamped case with solution (21) in section 3.4. In particular, the results of problems 3.4-32 and 3.4-33 apply. The quadratic formula: r = [-c/(2m)] + [(1/(2m)) sqrt(c^2 - 4km)] r = [-c/(2m)] - [(1/(2m)) sqrt(c^2 - 4km)] Because of the underdamped case, the argument of the square root is negative. The book defines in section 3.4 equations (21) to (23) symbols p and omega_1 via the equation r = [-p] + [omega_1 sqrt(-1)] r = [-p] - [omega_1 sqrt(-1)] Then m = 3.125 [given in 3.4-34], p = (1/2)(c/m), omega_1 = sqrt(4km-c^2)/(2m). Once you know both symbols p and omega_1, then you also know c and k. Equations (21) to (23) in section 3.4 describe the relations between m,c,k and the symbol omega_1 used in problems 32 and 33. The phase-amplitude form of the solution, as described in section 3.4, equations (12) to (14), is used to solve problems 32 and 33. Because problem 3.4-33 is solved in the student solution manual, you are able to reference details for that solution. No details are expected for the solution to 3.4-32, especially, do not include them with your solution to 3.4-34. Section 3.5 ===================== Updated S2015 3.5-6 2y'' + 4y' + 7y = x^2 ===== The characteristic equation 2r^2+4r+7=0 has roots r=-1+wi, r=-1-wi, w=sqrt(5). The atoms of the homogeneous equation are e^{-x}cos wx, e^{-x}sin wx. Because f(x)=x^2, the initial atom list for f(x) is x^2. Add all lower-power related atoms to obtain the list 1,x,x^2. Because none of the atoms 1,x,x^2 are solutions of the homogeneous equation, then no change is required. The corrected trial solution is y=d1 + d2 x + d3 x^2. Substitute the trial solution into the DE, solve for d1,d2,d3. The book's answer is correct. 3.5-12 y''' + y' = 2 - sin(x) ====== The characteristic equation is r^3 + r = 0. The roots are r=0,i,-i. The atom list for the homogeneous problem y''' + y' = 0 is 1,cos(x),sin(x). Because the right side of the differential equation is f(x)=2-sin(x), then the atoms in the trial solution are initially 1 and sin(x). Differentiation of this list of atoms leads to the modified list 1, cos(x), sin(x). Because 1, cos(x), sin(x) appear in the homogeneous solution, then the corrected trial solution uses the atom list x, x cos(x), x sin(x). Book: multiply atom 1 by x^s where s=1=multiplicity of the root r=0 in the characteristic equation r^3 + r = 0. Multiply atoms cos(x) and sin(x) by x^s where s=1=multiplicity of the root r=i in the characteristic equation r^3 + r = 0. Lectures: Identify three groups of related atoms: group 1: 1, group 2: cos(x), group 3: sin(x). For each group, cross out the homogeneous solution(s) and append the next atom in that group. The trial solution with fewest atoms is y = d1 (x) + d2 (x cos(x)) + d3 (x sin(x)). Substitute the trial solution into the DE, solve for the d-symbols. The book's answer is correct. 3.5-14 y'''' - 2y'' + y = x exp(x) ====== The characteristic equation is r^4-2r^2+1=0. The roots are r=1,-1,1,-1. The homogeneous DE y^(4) - 2y'' + y = 0 has atom list e^x, xe^x, e^{-x}, xe^{-x}. bBecause f(x)=xe^x, then the initial atom list for f(x) is xe^x. Differentiation expands the list to e^x, xe^x. This list forms a single group of related atoms. Because e^x and xe^x appear in the homogeneous solution, then both must be crossed-out from group 1, then atoms x^2e^x and x^3e^x are appended. The corrected trial solution is y=d1 (x^2e^x) + d2 (x^3e^x). Book: multiply the initial atom list by by x^s where s=2=multiplicity of the root r=1 in the characteristic equation r^4-2r^2+1=0. Lectures: Cross-out from group 1 both homogeneous solutions and add on the next atoms in the group, namely, x^2e^x and x^3e^x. Substitute the trial solution y into the DE, solve for the d-symbols. The book's answer is correct. 3.5-22 y^(5) - y''' = exp(x) + 2x^2 - 5 ====== The characteristic equation is r^5 - r^3 = 0. The roots are r=0,0,0,1,-1. The homogeneous equation y^(5) - y''' = 0 then has atom list 1, x, x^2, e^x, e^{-x}. Because f(x)=e^x + 2x^2 - 5, then the initial list of atoms is e^x, x^2, 1. Expand this list to 1, x, x^2, e^x [add all lower-power related atoms]. Because 1, x, x^2, e^x are atoms of the homogeneous solution, then the corrected atom list is x^3, x^4, x^5, xe^x Book: The trial solution is y = x^{s1}(d1 + d2 x + d3 x^2) + x^{s2}(d4 e^x) where s1=3 is the root multiplicity for root r=0 of r^5 - r^3 = 0, and s2=1 is the root multiplicity for root r=1 of r^5 - r^3 = 0. Lectures: Doing this construction from basic rules in class, we would alter the initial atom list as follows group 1: 1,x,x^2 Three homogeneous sols, cross out each, add x^3,x^4,x^5 group 2: e^x Homogeneous sol, cross it out, add xe^x Then the trial solution is y = d1 (x^3)+ d2 (x^4) + d3 (x^5) + d4 (x e^x) Substitute the [corrected] trial solution into the DE, solve for the d-symbols. The book's answer is correct. 3.5-27 y^(4) + 5y'' + 4y = sin(x) + cos(2x) ====== The characteristic equation is r^4+5r^2+4=0. The roots are complex, r=2i,-2i,i,-i. The homogeneous equation y^(4) + 5y'' + 4y = 0 then has atom list cos 2x, sin 2x, cos x, sin x. Because f(x)=sin x + cos 2x, then the initial list of atoms is sin(x), cos(2x). Expand this list to cos(x), sin(x), cos(2x), sin(2x). Because cos x and cos 2x are atoms of the homogeneous solution, then the corrected atom list is x cos(x), x sin(x), x cos(2x), x sin(2x). Book: The trial solution is y = x^{s1}(d1 cos x + d2 sin x) + x^{s2}(d3 cos 2x + d4 sin 2x) where s1=1 and s2=1 are root multiplicities for roots of r^4+5r^2+4=0. Lectures: Doing this construction from basic rules in class, we would alter the initial atom list as follows group 1: cos x Homogeneous sol, cross it out, add (x cos x) group 2: sin x Homogeneous sol, cross it out, add (x sin x) group 3: cos 2x Homogeneous sol, cross it out, add (x cos 2x) group 4: sin 2x Homogeneous sol, cross it out, add (x sin 2x) Then the trial solution is y = d1 (x cos x)+ d2 (x sin x) + d3 (x cos 2x) + d4 (x sin 2x) Substitute the [corrected] trial solution into the DE, solve for the d-symbols. The book's answer is correct. 3.5-48 y'' - 2y' - 8y = 3 exp(-2x) ====== There are two terms in the hand-generated solution, which uses formula (33) of 3.3. One looks like y_h(x). Removing it leaves the shortest expression for y_p(x): -(x/2)exp(-2x). The answer can be tested from the maple function odetest() as follows: ODE := diff(y(x),x,x)-2*diff(y(x),x)-8*y(x)=3*exp(-2*x); sol := y(x)=(-1/2)*x*exp(-2*x); odetest(sol,ODE,y(x)); 3.5-50 y'' - 4y = sinh(2x) ====== DEF: sinh(u) = (1/2)(exp(u)-exp(-u)) The atoms for characteristic equation r^2-4=0 are e^{2x}, e^{-2x}. Use these for y1, y2 in the method of variation of parameters, formula (33) in the textbook section 3.3. Then W=Wronskian(y1,y2)= y1 y2' - y1' y2 = -2 - 2 = -4, f(x)=RHS of the DE = sinh 2x = (1/2)(e^{2x}-e^{-2x}) yp=C1 y1 + C2 y2 [formula (33) in section 3.5] C1 = int(-f y2/W)=int((1/8)-(1/8)e^{-4x})=x/8 + e^{-4x}/32 C2 = int( f y1/W)=int((-1/8)e^{4x}+(1/8))=x/8 - e^{4x}/32 yp=(1/8)xe^{2x}+(1/8)xe^{-2x}+(1/32)e^{-2x}-(1/32)e^{2x} Drop the yh terms to get the shortest solution yp=(1/8)xe^{2x}+(1/8)xe^{-2x}+(yh terms to be dropped) yp=(1/8)xe^{2x}+(1/8)xe^{-2x} 3.5-52 y'' + 9y = sin(3x) ====== The answer obtained by integration by hand contains three terms, two of which collect to (1/18)sin(3x)(sin^2(3x)+cos^2(3x)) and reduce to (1/18)sin(3x), a special case of y_h(x). Since y_h(x) should not appear in the shortest possible y_p(x), we remove it to obtain a single term for the answer: -(x/6)cos(3x). 3.5-54 y'' + y = csc^2(x) ====== DEF: csc(u)=1/sin(u), sec(u)=1/cos(u), tan(u)=sin(u)/cos(u), cot(u)=1/tan(u) EQUATIONS: cos^2(u)+sin^2(u)=1, 1+tan^2(u)=sec^2(u), cot^2(u)+1=csc^2(u) The atoms for characteristic equation r^2+1=0 are cos(x), xin(x). Use these for y1, y2 in the method of variation of parameters, formula (33) in the textbook section 3.3. Then W=Wronskian(y1,y2)= y1 y2' - y1' y2 = cos^2 x + sin^2 x = 1, f(x)=RHS of the DE = csc^2(x), yp=C1 y1 + C2 y2 [formula (33) in section 3.5] C1 = int(-f y2/W)=int(-csc^2(x)sin(x),x)=-ln|csc(x)-cot(x)|, C2 = int( f y1/W)=int( csc^2(x)cos(x),x)=int(csc(x)cot(x),x)=-csc(x), yp=-cos(x)ln|csc(x)-cot(x)| - csc(x)sin(x) Drop the yh terms to get the shortest solution yp=-cos(x)ln|csc(x)-cot(x)| - 1 3.5-58 x^2y'' - 4xy' + 6y = x^3; y_h = c_1 x^2 + c_2 x^3 ====== The standard form of the DE, required to directly use formula (33) in this section of the book, is y'' +(-4/x)y'+ (6/x^2)y=x The functions y1, y2 in (33) are a basis for the solution space of the homogeneous DE, which is given to have general solution yh = c1 (x^2) + c2 (x^3). Take y1, y2 to be the partials on symbols c1, c2: y1 = x^2, y2 = x^3 Then W = y1 y2' - y1' y2 = x^2(3x^2)-x(2x)(x^3) = x^4 f(x)=x^3/x^2=x [RHS of DE in standard form] yp = C1 y1 + C2 y2 [variation of parameters (33)] C1 = int(-f y2/W)=int(-x(x^3)/x^4,x)=-x, C2 = int(f y1/W)=int(x(x^2)/x^4,x)=ln|x| yp = C1 y1 + C2 y2 = -x^3 + x^3 ln|x| 3.5-60 4x^2y'' - 4xy' + 3y = 8 x^(4/3); y_h = c_1 x + c_2 x^(3/4) ====== The problem statement has a typo but the book's answer in BOB is correct. The formula given in the problem statement should be y_c = c1 (x^(1/2)) + c2 (x^(3/2)) Use y1:=sqrt(x); and y2:=x*sqrt(x); to evaluate integrals in the variation of parameters formula yp := y1*int(y2*(-f(x))/w,x) + y2*int(y1*f(x)/w,x); where w:=y1*diff(y2,x)-y2*diff(y1,x); Hint from Jon Engle, 10:45 class. If you got a coefficient of 72/7, then likely you used the formula for y_c given in the problem, which means you should change y_c to the one above. 3.5-62 (x^2 - 1)y'' - 2xy' + 2y = x^2 - 1; y_h=c_1 x + c_2(1+x^2) ====== The standard form of the DE, required to directly use formula (33) in the textbook section, is y'' +((-2x/(x^2-1))y'+ (2/(x^2-1))y=1 The functions y1, y2 in (33) are a basis for the solution space of the homogeneous DE, which is given to have general solution yh = c1 x + c2 (1+x^2). Take y1, y2 to be the partials on symbols c1, c2: y1 = x, y2 = 1+x^2 Then W = y1 y2' - y1' y2 = 2x^2-1-x^2 = x^2-1 f(x)=(x^2-1)/(x^2-1)=1 [RHS of DE in standard form] yp = C1 y1 + C2 y2 [variation of parameters (33)] C1 = int(-f y2/W)=-x-ln|x-1|+ln|x+1| C2 = int(f y1/W)=(1/2)ln|x-1|+(1/2)ln|x+1| Section 3.6 ===================== Updated S2015 3.6-2 x'' + 4x = 10 cos(2t), x(0)=x'(0)=0 ===== 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. 3.6-4 x'' + 25x = 90 cos(4t), x(0)=0, x'(0)=90 ===== x'' + 25 x = 90 cos 4t, x(0)=0, x'(0)=90 This is a BEATS problem. For another example, and more details, see the problem 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. 3.6-8 x'' + 3x' + 5x = -4 cos 5t ===== Find the steady-state periodic solution. Graph it and F(t)/(m omega) on one axes. 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 3.6 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 4 of the chapter, especially between equations (11)-(12), to find out what to do. Beware of the advice in section 6, which suggests the calculator answer or the calculator answer plus Pi [see (22) in section 6], because that advice applies only to the special problem being studied in section 6. GRAPH. Because m=1 and omega=5, then F(t)/(m omega) = (-4/5) cos(5t). Already you have x(t) = C cos(5t - alpha). Graph the two functions on one set of axes. For graphing, you can also use the alternate form x(t) = (-A) cos 5t + (-B) sin 5t, which may help you to detect an error in computing alpha. 3.6-10 x'' + 3x' + 3x = 8 cos(10t) + 6 sin(10t) ===== Find the steady-state periodic solution. Graph it and F(t)/(m omega) on one axes. 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 3.6-12 x'' + 6 x' + 13 x = 10 sin(5t), x(0)=0, x'(0)=0 ====== Find the solution and the steady-state solution. Plot them on a single set of axes. 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 3.6. 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 3.4 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. 3.6-16 x'' + 4x' + 5x = 10 cos(wt). Find a formula for C(w). ====== Graph C(w) against w. Test for practical resonance. 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 3.6-27] 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 3.6-18, in which w=sqrt(600) and practical resonance does occur. 3.6-18 x'' + 10x' + 650x = 100 cos(wt). Find a formula for C(w). ====== Graph C(w) against w. Test for practical resonance. 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 section 6, problem 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. AMPLITUDE C(w). Use formula (21) in Edwards-Penney section 6. It says for the equation x'' + 10x' + 650x = 100 cos(wt) that C(w) = 100/sqrt((650-w^2)^2 + (10w)^2) Graph C against w with a calculator or maple, and trace a replica of the graph into your solution. Section 3.7 ===================== Updated 2015 3.7-4 LRC-circuit, L=2, R=40, E=100 exp(-10t); I(0)=0 The RL-circuit equation for the current I(t) is L I'(t) + R I(t) = E(t) Then L=2, R=40, E=100 exp(-10 t), initial current zero, gives 2 I'(t) + 40 I(t) = 100 exp(-10 t), I(0)=0 which in standard form is I'(t) + 20 I(t) = 50 exp(-10 t), I(0)=0 =============== CHAPTER 1 METHOD. This is a first order linear differential equation with constant coefficients, first treated in section 1.5 of Edwards-Penney. The answer from the superposition principle is I(t) = I_h + I_p where I_h = homogeneous solution = constant/integrating factor. A particular solution I_p can be found by undetermined coefficients. =========================== LAPLACE METHOD (RECOMMENDED). The isolated equation for L(I(t)) is 50 L(I(t)) = ------------ = Transfer x L(input) (s+20)(s+10) Then partial fractions and backward table methods give the solution I(t) = 5 exp(-10 t) - 5 exp(-20 t) MAXIMUM. To find the maximum on t >= 0, differentiate the formula for I(t), set it equal to zero, and solve for t. Substitute this special value of t back into the formula for I(t), to obtain the maximum current I. ANSWER CHECK. The maple answer check appears below. Symbol % means the answer from the previous line. de:=2*diff(i(t),t)+40*i(t)=100*exp(-10*t); ic:=i(0)=0; dsolve({de,ic},i(t)); u:=unapply(rhs(%),t); diff(u(t),t); # find the derivative of the current solve(%=0,t); # set derivative equal to zero and solve for t u(%); # evaluate current at this value of t ================================================================ 3.7-12 R=200, L=5, C=1/1000; E=100 sin(10t) The equation to be solved is 5 Q'' + 30 Q' + 50 Q = 50 sin(2 t) 5 I'' + 30 I' + 50 I = 100 cos(2t) [diff prev eq] The steady-state solution is the undetermined coefficients solution I(t) of the current equation, I(t) = d1 cos(2t) + d2 sin(2t) Using formulas in the textbook is easier, and this gives rise to the phase-amplitude equation directly: I(t) = (E0 / Z) cos(2t - alpha) E0 = 50 Z = impedance from formula (11) in the text alpha = arctan(...) from formula (9) in the text === end Chapter 3 ===