MATH 285 E1 F1 GRADED HOMEWORK SET 3 DUE WEDNESDAY OCTOBER 8 IN LECTURE This time the homework has just one part Please staple your homework together and put your name and section on it Thank you 1 5 points Show that the functions e3x xe3x and x2 e3x are linearly independent That is show that if c1 e3x c2 xe3x c3 x2 e3x 0 for all x then it necessarily follows that c1 c2 c3 0 Hint Get relations between the ci s by plugging in x 0 and or differentiating the equation Plug in x 0 to get c1 e0 c2 0e0 c3 0e0 0 Thus c1 0 With that knowledge we can differentiate to get c2 3xe3x e3x c3 3x2 e3x 2xe3x 0 Plug in x 0 to get c2 0 e0 c3 0 0 0 Thus c2 0 With that knowledge we can differentiate a second time to get c3 9x2 e3x 6xe3x 6xe3x 2e3x 0 Plug in x 0 to get c3 0 0 0 2e0 0 Thus c3 0 and we are done 2 15 points Consider the polynomial p r r12 12r11 51r10 r9 1968r8 19003r7 106948r6 440432r5 1423168r4 3448064r3 6069248r2 7606272r 5160960 Find the general solution of the linear homogeneous equation p D y 0 d and p D denotes the constant coefficient where as usual D dx differential operator obtained by substituting D for r in p r We are looking for a general solution that is real not complex and which involves twelve arbitrary constants 1 2 GRADED HOMEWORK 3 Hint You should use the following factorization of p r p r r 3 2 r 4 3 r 7 r2 r 5 r2 16 2 The given factorization of p r gives a factorizaton of p D p D D 3 2 D 4 3 D 7 D2 D 5 D2 16 2 The general solution of p D y 0 is a combination of the general solutions for each of the factors We proceed factor by factor a D 3 2 y 0 There is a real double root at 3 The basic solutions are e3x and xe3x The general solution is y c1 e3x c2 xe3x b D 4 3 y 0 There is a real triple root at 4 The basic solutions are e4x xe4x and x2 e4x The general solution is y c3 e4x c4 xe4x c5 x2 e4x c D 7 y 0 There is a real simple root at 7 The basic solution is e 7x The general solution is y c6 e 7x d D2 D 5 y 0 There are complex roots at 19 1 i 2 2 There is a complex solution 19 1 2 i 19 2 x 1 2 x i 19 2 x 1 2 x e e e e cos x i sin 2 19 x 2 The real and imaginary parts of this solution are also solutions so we have basic solutions 19 19 1 2 x 1 2 x e cos x e sin x 2 2 The general solution is 19 y c7 e 1 2 x cos x c8 e 1 2 x sin 2 19 x 2 e D2 16 2 The characteristic polynomial r2 16 2 factors into complex factors as r2 16 2 r 4i r 4i 2 r 4i 2 r 4i 2 Thus there are double roots at 4i Thus there are complex solutions e4ix and e 4ix and additionally xe4ix and xe 4ix The real and imaginary parts of the solution e4ix cos 4x i sin 4x GRADED HOMEWORK 3 3 namely cos 4x and sin 4x are also solutions of the equation as are the real and imaginary parts of xe4ix x cos 4x ix sin 4x Thus the four basic real solutions are cos 4x sin 4x x cos 4x and x sin 4x The general solution is y c9 cos 4x c10 sin 4x c11 x cos 4x c12 x sin 4x Putting it all together the general solution of the equation p D y 0 is y c1 e3x c2 xe3x c3 e4x c4 xe4x c5 x2 e4x c6 e 7x c7 e 1 2 x cos 19 x 2 c8 e 1 2 x sin 19 x 2 c9 cos 4x c10 sin 4x c11 x cos 4x c12 x sin 4x Or in a slightly more compact form y c1 c2 x e3x c3 c4 x c5 x2 e4x 19 7x 1 2 x x c8 sin c6 e e c7 cos 2 19 x 2 c9 cos 4x c10 sin 4x x c11 cos 4x c12 sin 4x 3 10 points A body of mass m 5 kg is attached to a spring with spring constant k 20 kg s2 The body is suspended on an airhockey table so that it is not subject to friction or gravity but it is connected to a dashpot mechanism that lets us adjust the degree of damping c 0 The equation of motion is the usual one d2 x dx c kx 0 2 dt dt Consider the following experimental procedure move the body to an initial position x 0 0 m and set it into motion with initial velocity v 0 1 m s Your task is to determine the position function x t in the following three cases a c 10 kg s b c 20 kg s c c 30 kg s It is perfectly fine to refer to section 3 4 of the textbook in your solution that is you don t have to rederive everything all over again but you should mention what facts from the textbook you are using We will use SI units througout The characteristic equation is 2 5r cr 20 0 Its discriminant is c2 4 5 20 c2 400 The nature of the solutions depends on whether c2 400 c2 400 or m 4 GRADED HOMEWORK 3 c2 400 The first is underdamped the second is critically damped and the third is overdamped a c 10 Thus c2 400 300 0 so this is underdamped The roots of the characteristic equation are 10 300 r 1 i 3 10 The general solution is given by equation 21 in section 3 4 x t e t A cos 3t B sin 3t Using the initial condition x 0 0 we get 0 e0 A cos 0 B sin 0 A Thus A 0 Differentiating the solution we get v t e t B sin 3t 3e t B cos 3t Using the initial condition v 0 1 we get 1 3B or B 1 3 So the solution is 1 x t e t sin 3t 3 b c 20 Thus c2 400 0 so this is critically damped The characteristic equation has just one root r 20 2 10 The general solution is given by equation 20 in section 3 4 x t e 2t c1 c2 t Using the initial condition x 0 0 we get 0 e0 c1 c2 0 Thus c1 0 Differentiating the solution we get v t 2e 2t c2 t e 2t c2 Using v 0 …
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