1. Homework 11.1. Linear diffeq's and recursions1. Homework 1 answers1.1. Linear diffeq's and recursionsMATH 56A: FALL 2006HOMEWORK AND ANSWERS1. Homework 11.1. Linear diffeq’s and recursions. Three problems:0.1 Find all functions x(t), y(t) so thatx0(t) = −x + y, y0(t) = 3x − 3yFind the particular solution so that x(0) = y(0) = 1/2.0.5 Find all functions f from integers to real numbers so thatf(n) =12f(n + 1) +12f(n − 1) − 1[Show first that f(n) = n2is a particular solution.]0.6 (a) Find all functions f : R → R so thatf00(x) + f0(x) + f(x) = 0(b) Final all functions f : Z → R so thatf(n + 2) + f(n + 1) + f(n) = 012 HW AND ANSWERS 20061. Homework 1 answers1.1. Linear diffeq’s and recursions. four answers:0.1 Find all functions x(t), y(t) so thatx0(t) = −x + y, y0(t) = 3x − 3yFind the particular solution so that x(0) = y(0) = 1/2.The matrix isA =−1 13 −3This has eigenvalues 0, −4 with corresponding eigenvectors X1=11, X2=1−3. So A =QDQ−1whereQ =1 11 −3, D =0 00 −4, Q−1=143 11 −1AndetA= QetDQ−1=141 11 −31 00 e−4t3 11 −1=143 + e−4t1 − e−4t3 − 3e−4t1 + 3e−4tThe general solution is X = etAX0orx =x043 + e−4t+y041 − e−4ty =x043 − 3e−4t+y041 + 3e−4tWhen x0= y0then the e−4tterms all cancel and we get that x, y are constant functions. Inparticular, x = y =12is the particular solution in the homework.Some people found another method but didn’t carry it through: Some of you noticed thatthe equations say: y0= −3x0ordydt= −3dxdtCancel the dt’s and integrate:Zdy =Z−3dxwhich gives: y = −3x + C1. Now you have to continue and put it back into the originalequationdxdt= −x + y = −x − 3x + C1= −4x + C1dx−4x + C1= dt−14ln | − 4x + C1| = t + C2HW AND ANSWERS 2006 3−4x + C1= ±e−4t−4C2So,x = C3e−4t+14C1andy = −3x + C1= −3C3e−4t+14C1When you put in the initial conditions you find C1= 2, C3= 0.Remember that you need to add +C with a new C every time you integrate.0.5 Find all functions f from integers to real numbers so thatf(n) =12f(n + 1) +12f(n − 1) − 1[Show first that f(n) = n2is a particular solution.]To solve the homogeneous equation, try f = an. If a is a double root then the secondsolution is f(n) = nan. The homogenous equation givesa2− 2a + 1 = 0This has only one root: a = 1. So, the solutions are f (n) = 1 and f(n) = n. Thus thegeneral solution isn2+ bn + cwhere b and c are constants. There is no constant in front of the particular solution.0.6 (a) Find all functions f : R → R so thatf00(x) + f0(x) + f(x) = 0Here you try f (x) = eλxand you find thatλ2+ λ + 1 = 0orλ =−1 ± i√32eλx= e−x/2(cos√3x2± i sin√3x2)To get a real solution students correctly took a linear combination of the real and imaginaryparts:f(x) = ae−x/2cos√3x2+ be−x/2sin√3x2(b) Find all functions f : Z → R so thatf(n + 2) + f(n + 1) + f(n) = 04 HW AND ANSWERS 2006You try f (n) = anand you geta2+ a + 1 = 0Ora =−1 ± i√32So, an arbitrary complex solution is given byf(n) = a −1 + i√32!n+ b −1 − i√32!nIn order for this to be a real number it must be equal to its complex conjugate. So, b = a.I.e., a = c + id, b = c − id wherec = f (0)/2d = −f
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