c 1 = y 0, c 2 = 1 2 y 1. Therefore the solution to the general initial-value problem is y(t) = y 0 cos(2t)+y sin(2t).

Transcription

1 Solutions to Second In-Class Exam Math 246, Professor David Levermore Tuesday, 29 October 2 ( [4] Give the interval of definition for the solution of the initial-value problem u t u + cos(5t 6+t u = et 2 t, u( = u ( = u ( =. Solution. The equation is linear and is already in normal form. The coefficient of u is undefined at t = and is continuous elsewhere. The coefficient of u is undefined at t = 6 and is continuous elsewhere. The forcing is undefined at t = 2 and is continuous elsewhere. Therefore the interval of definition is ( 6, because the initial time t = is in the interval ( 6,, all the coefficients and the forcing are continuous over the interval ( 6,, at t = 6 the coefficient of u is undefined, at t = the coefficient of u is undefined. (2 []Thefunctionscos(2tandsin(2tareafundamentalsetofsolutionstoy +4y =. (a Find the solution Y(t to the general initial-value problem y +4y =, y( = y, y ( = y. (b Find the associated natural fundamental set of solutions to y +4y =. Solution (a. Because we are given that cos(2t and sin(2t is a fundamental set of solutions to y +4y =, a general solution is y(t = c cos(2t+c 2 sin(2t. Because y (t = 2c sin(2t+2c 2 cos(2t, the initial conditions imply We solve these equations to obtain y = y( = c, y = y ( = 2c 2. c = y, c 2 = 2 y. Therefore the solution to the general initial-value problem is y(t = y cos(2t+y 2 sin(2t. Solution (b. We see from the above solution to the general initial-value problem that the associated natural fundamental set of solutions is N (t = cos(2t, N (t = 2 sin(2t. ( [4] Suppose that X (t, X 2 (t, and X (t are solutions of the differential equation x +2x t x +e t x =, Suppose you know that W[X,X 2,X ]( = 5. What is W[X,X 2,X ](t? Solution. Abel s Theorem states that w(t = W[X,X 2,X ](t satisfies w +2w =. It follows that w(t = w(e 2t. Because w( = W[X,X 2,X ]( = 5, we obtain w(t = 5e 2t. Therefore W[X,X 2,X ](t = 5e 2t.

2 2 (4 [] Give a general real solution of the equation D 2 x+5dx 6x = 2cos(2t, where D = d dt. Solution. This is a nonhomogeneous linear equation with constant coefficients. Its characteristic polynomial is p(z = z 2 +5z 6 = (z +6(z, which has the two simple real roots 6 and. Therefore a general solution of the associated homogeneous equation is x H (t = c e 6t +c 2 e t. The forcing 2 cos(2t has characteristic form with degree d = and characteristic µ+iν = i2, which is a root of p(z of multiplicity m =. Therefore we can use either Key Identity Evaluations or Undetermined Coefficients to find a particular solution x P (t. Then a general solution will be given by x(t = x H (t+x P (t. Key Indentity Evaluations. Because m = and m + d = for the forcing, we need only the Key Identity By evaluating this at z = i2 we obtain L ( e zt = p(ze zt = (z 2 +5z 6e zt. L ( e i2t = ( (i2 2 +5(i2 6 e i2t = ( 4+i 6e it = ( ie i2t. Because the forcing is 2cos(2t = 2Re(e i2t, we write L ( 2ei2t = 2e i2t, i which implies that ( ( ( 2e i2t e i2t +i (+ie i2t x P (t = Re = 2Re = 2Re i i +i = Re ( (+i(cos(2t+isin(2t = ( cos(2t sin(2t = cos(2t+sin(2t. Therefore a general solution of the equation is x(t = c e 6t +c 2 e t cos(2t+sin(2t. Undetermined Coefficients. Because µ + iν = i2 and m + d = m = for the forcing 2 cos(2t, we seek a particular solution in the form Because x P (t = Acos(2t+Bsin(2t. x P (t = 2Asin(2t+2Bcos(2t, x P (t = 4Acos(2t 4Bsin(2t,

3 we see that Lx P (t = x P(t+5x P(t 6x P (t = [ 4Acos(2t 4Bsin(2t ] +5 [ 2Asin(2t+2Bcos(2t ] 6 [ Acos(2t+Bsin(2t ] = ( A+Bcos(2t+( A Bsin(2t. By setting Lx P (t = 2cos(2t, we see that A+B = 2, A B =. The second equation implies A = B, which when placed into the first equation yields 2B = 2. Hence, B = and A =, which gives x P (t = cos(2t+sin(2t. Therefore a general solution of the equation is x(t = c e 6t +c 2 e t cos(2t+sin(2t. (5 [] What answer will be produced by the following Matlab commands? >> ode = D2y 6*Dy + *y = 5*exp(7*t ; >> dsolve(ode, t ans = Solution. The commands ask Matlab to give a general solution of the equation D 2 y 6Dy+y = 5e 7t, where D = d dt. This is a nonhomogeneous linear equation with constant coefficients. Its characteristic polynomial is p(z = z 2 6z + = (z 2 +4 = (z , which has the conjugate pair of roots ± i2. Therefore a general solution of the associated homogeneous equation is y H (t = c e t cos(2t+c 2 e t sin(2t. The forcing 5e 7t has characteristic form with degree d = and characteristic µ+iν = 7, which is a root of p(z of multiplicity m =. Therefore we can use either Key Identity Evaluations or Undetermined Coefficients to find a particular solution y P (t. Key Indentity Evaluations. Because m + d = m = we only need the Key identity, which is By evaluating this at z = 7 we obtain L ( e zt = p(ze zt = (z 2 6z +e zt. L ( e 7t = ( e 7t = (49 42+e 7t = 2e 7t. Becausetheforcing is5e 7t, we seethat aparticularsolutionisy P (t = 4 e7t. Therefore a general solution is y(t = c e t cos(2t+c 2 e t sin(2t+ 4 e7t.

4 4 Undetermined Coefficients. Because µ + iν = 7 and m + d = m = for the forcing 5e 7t, we seek a particular solution in the form y P (t = Ae 7t. Because y P (t = 7Ae7t and y P (t = 49Ae7t, we see that Ly P (t = y P(t 6y P(t+y P (t = [ 49Ae 7t] 6 [ 7Ae 7t] + [ Ae 7t] = (49 42+Ae 7t = 2Ae 7t. By setting Ly P (t = 5e 7t, we see that 2A = 5, whereby A =. Hence, we obtain 4 the particular solution y P (t = 4 e7t. Therefore a general solution is y(t = c e t cos(2t+c 2 e t sin(2t+ 4 e7t. (6 [8] Compute the Green function g(t associated with the differential operator L = D 2 4D+4, where D = d dt. Solution. Because the differential operator L has constant coefficients, the Green function g(t associated with it satisfies the initial-value problem Lg = D 2 g 4Dg +4g =, g( =, g ( =. The characteristic polynomial is p(z = z 2 4z +4 = (z 2 2, which has the double real root 2. Hence, the Green function has the form g(t = c e 2t +c 2 te 2t. The initial condition g( = implies that c =. Because g (t = 2c 2 te 2t +c 2 e 2t, the initial condition g ( = implies that c 2 =. Therefore the Green function is (7 [8] Solve the initial-value problem g(t = te 2t. w 4w +4w = 2e2t +t 2, w( = w ( =. Solution. BythepreviousproblemtheGreenfunctionforthisproblemisg(t = te 2t. Because this equation is in normal form, the solution to this inital-value problem is given by the Green function formula w(t = = 2te 2t g(t sf(sds = +s 2 ds e2t (t se 2(t s 2e 2s +s ds 2 2s +s 2 ds.

5 Because we find that +s ds = 2 tan (s t = tan (t, s= 2s +s ds = 2 log(+s2 t = log(+t 2, s= w(t = 2te 2t tan (t e 2t log(+t 2. Remark. This problem can also be solved by using variation of parameters. However that approach is not as efficient because it does not directly solve the initial-value problem. Rather, after finding a particular solution the constants c and c 2 in Y H (t must be determined to satisfy the initial conditions. (8 [8] Find a particular solution v P (t of the equation v 9v = 2e t. Solution. This is a nonhomogeneous linear equation with constant coefficients. Its characteristic polynomial is p(z = z 2 9 = (z +(z, which has two simple real roots and. The forcing 2e t has characteristic form with degree d = and characteristic µ+iν =, which is a root of p(z of multiplicity m =. Therefore we can use either Key Identity Evaluations or Undetermined Coefficients to find a particular solution x P (t. Key Indentity Evaluations. Because m = and m+d = for the forcing, we only need the first derivative of the Key Identity. The Key Identity and its first derivative are L ( e zt = (z 2 9e zt, L ( te zt = (z 2 9te zt +2ze zt. By evaluating the derivative of the Key identity at z = we obtain L ( te t = 2 e t = 6e t. Because the forcing is 2e t, we see that a particular solution is x P (t = 2te t. Undetermined Coefficients. Because µ + iν = and m + d = m = for the forcing 2e t, we seek a particular solution in the form Because we obtain x P (t = Ate t. x P (t = Atet +Ae t, x P (t = 9Atet +6Ae t, Lx P (t = [ 9Ate t +6Ae t] 9 [ Ate t] = 6Ae t. By setting Lx P (t = 2e t, we see that 6A = 2, whereby A = 2. Therefore, a particular solution is x P (t = 2te t. Remark. A general solution is x(t = c e t +c 2 e t +2te t. 5

6 6 (9 [] Let L be a linear ordinary differential operator with constant coefficients. Suppose that all the roots of its characteristic polynomial (listed with their multiplicities are +i2, +i2, i2, i2, i5, i5, 4, 4,,. (a Give the order of L. (b Give a general real solution of the homogeneous equation Ly =. Solution (a. There are roots listed above, so the degree of the characteristic polynomial is, whereby the order of L is also. Solution (b. A general solution is y(t = c e t cos(2t+c 2 e t sin(2t+c te t cos(2t+c 4 te t sin(2t +c 5 cos(5t+c 6 sin(5t+c 7 e 4t +c 8 te 4t +c 9 +c t. Here the fundamental set of solutions is generated as follows: the double conjugate pair ±i2 yields e t cos(2t, e t sin(2t, te t cos(2t, and te t sin(2t; the single conjugate pair ±i5 yields cos(5t and sin(5t; the double real root 4 yields e 4t and te 4t ; the double real root yields and t. ( [] The functions +t and e t are solutions of the homogeneous equation ty (+ty +y = over t >. (You do not have to check that this is true! (a Show that these functions are linearly independent. (b Give a general solution of the nonhomogeneous equation ty (+ty +y = t2 +t over t >. Solution (a. The Wronskian of +t and e t is W [ ( +t,e t] +t e t (t = det e t = (+t e t e t = e t +te t e t = te t. Because W[ + t,e t ](t = te t for t >, the functions + t and e t are linearly independent. Solution (b. Because +t and e t are linearly independent, a general solution of the associated homogeneous problem is y H (t = c (+t+c 2 e t. Because this problem has variable coefficients, you should use either the general Green Function method or Variation of Parameters to find a particular solution y P (t. Both of these methods require the equation to be put into its normal form, which is y +t t y + t y = t +t.

7 7 General Green Function. The Green function G(t, s is given by ( +s e G(t,s = W[+s,e s ](s det s +t e t = (+set (+te s. se s The Green function formula with t I = then yields the solution y P (t = G(t,sf(sds = = e t e s ds+(+t We can evaluate the above definite integrals as e s ds = e s t s= +s ds = log(+s t (+se t (+te s se s +s ds. s +s ds = e e t, ( +t = log(+t log(2 = log. s= 2 Therefore a general solution is ( +t y(t = c (+t+c 2 e t e t ++(+tlog. 2 Variation of Parameters. We seek a solution in the form y(t = u (t(+t+u 2 (te t. where u (t and u 2 (t satisfy the linear algebraic system u (t(+t+u 2 (tet =, The solution of this system is u (t+u 2 (tet = t +t. u (t = +t, u 2 (t = e t. Alternatively, because W[+t,e t ](t = te t, the formulas from the notes yield u (t = et te t ( t +t = +t, u 2(t = +t te t ( t +t = e t. No matter how they are obtained, you integrate these equations to find u (t = +t dt = c +log(+t, u 2 (t = e t dt = c 2 +e t. Therefore a general solution is y(t = ( c +log(+t (+t+ ( c 2 +e t e t = c (+t+c 2 e t ++(+tlog(+t. Remark. This general solution appears different than the one we obtained by the general Green function method. However, by replacing c and c in this solution with c log(2 and c 2 e it transforms into the earlier one, so they are equivalent.

8 8 ( [] The vertical displacement of an unforced mass on a spring is given by h(t = cos(2t 4sin(2t. (a Is this system undamped, under damped, critically damped, or over damped? (Give your reasoning! (b Express h(t in the amplitude-phase form h(t = Acos(2t δ with A > and δ < 2π. Label the amplitude and phase. (The phase may be expressed in terms of an inverse trig function. (c Give the natural frequency and period of this spring-mass system. (d If an external force F ext (t = 7sin(ωt is applied, at what value of the driving frequency ω does resonance occur? Solution (a. The system is undamped because the given displacement h(t comes from an underlying characteristic polynomial having the imaginary conjugate pair of roots ±i2. Solution (b. By comparing Acos(2t δ = Acos(δcos(2t+Asin(δsin(2t, with h(t = cos(2t 4sin(2t, we see that Acos(δ =, Asin(δ = 4. This shows that (A,δ are the polar coordinates of the point in the plane whose Cartesian coordinates are (, 4. Clearly A is given by A = 2 +( 4 2 = 9+6 = 25 = 5. Because (, 4 lies in the fourth quadrant, the phase δ satisfies π < δ < 2π. We 2 can express δ several ways. A picture shows that if we use 2π as a reference then sin(2π δ = 4 5, cos(2π δ = 5, tan(2π δ = 4, whereby we can express the phase by any one of the formulas δ = 2π sin ( ( ( 4 5, δ = 2π cos 5, δ = 2π tan 4. The same picture shows that if we use π as a reference then 2 sin(δ π =, cos(δ π = 4, tan(δ π =, whereby we can express the phase by any one of the formulas δ = π +sin ( 2 5, δ = π +cos ( 4 2 5, δ = π +tan ( 2 4. Only one expression for δ is required. Remark. It is incorrect to give the phase by one of the formulas δ = sin ( 5 ( 4, δ = cos 5 (, δ = tan 4, because, by our conventions for the range of the inverse trigonometric functions, sin ( 4 5 lies in ( π 2,, cos ( 5 lies in (, π, while tan ( 4 2 lies in (, π 2. Solution (c. The natural frequency ω o is 2 radians per sec, so the natural period T o is given by T o = 2π ω o = 2π 2 = π sec.

9 Solution (d. Resonance occurs when the driving frequency ω equals the natural frequency ω o. Given the answer to part (c, resonance occurs when ω = ω o = 2 radians per sec. (2 [8] When a gram mass is hung vertically from a spring, at rest it stretches the spring 4.9 cm. (Gravitational acceleration is g = 98 cm/sec 2. At t = the mass is displaced 5 cm above its rest position and is released with a downward velocity of 2 cm/sec. The medium imparts a damping force of 9 dynes ( dyne = gram cm/sec 2 when the speed of the mass is cm/sec. Assume that the spring force is proportional to displacement, that the damping is proportional to veloicity, and that there are no other forces. (a Formulateaninitial-valueproblemthat governs themotionofthemass fort >. (DO NOT solve this initial-value problem, just write it down! (b Is this system undamped, under damped, critically damped, or over damped? (Give your reasoning! Solution (a. Let h(t be the displacement (in centimeters of the mass from its rest position at time t (in seconds, with upward displacements being positive. The governing initial-value problem then has the form mh +γh +kh =, h( = 5, h ( = 2, where m is the mass, γ is the damping coefficient, and k is the spring constant. We are given that m = grams. We obtain k by balancing the force applied by the spring when it is stetched 4.9 cm with the weight of the mass (mg = 98 dynes. This gives k4.9 = 98, or k = 98 = 2 = 2 dynes/cm. 4.9 We obtain γ by balancing the damping force when the speed of the mass is cm/sec with 9 dynes. This gives γ = 9, or γ = 9 = dynes sec/cm Therefore the governing initial-value problem is h +h +2h, h( = 5, h ( = 2. Remark. Had we chosen the convention of downward displacements being positive then the governing initial-value problem is h +h +2h, h( = 5, h ( = 2. Solution (b. The normal form of the governing equation is Its charactericitc polynomial is h +h +2h. p(z = z 2 +z +2 = (z +(z +2. This has real roots and 2. Therefore the system is over damped. 9