Solutions to Midterm #2 Practice Problems

Transcription

1 Solutions to Midterm # Practice Problems 1. Find general solutions to the following DEs: a) y 6y + 8y = 0 Solution: The characteristic equation for this constant-coefficient, homogeneous DE is r 6r + 8 = 0, which factors as r )r ) = 0. Therefore, its roots are r = and r =, so the general solution is y = c 1 e x + c e x. b) y ) + y y 8y = 0 Solution: The characteristic equation for this constant-coefficient, homogeneous DE is r + r r 8 = 0. We see immediately that r + is a factor of the polynomial; factoring it out, we have r + )r ) = r + )r + )r ) = 0. Therefore, r = is a single root, and r = is a double root, so the general solution is y = c 1 e x + c e x + c xe x. c) y + 8y + 0y = 0 Solution: The characteristic equation is r + 8r + 0 = 0, which has no immediately obvious solutions. Using the quadratic formula, r = 8 ± 6 80 = 8 ± 16 = 8 ± i = ± i. From this pair of complex roots, the general solution is y = c 1 e x cos x + c e x sin x. d) y ) = y Solution: Writing this DE as y ) y = 0, its chararcteristic equation is r 1 = 0. This factors as r 1)r + 1)r + 1) = 0, so it has roots r = 1, r = 1, and r = ±i. Therefore, the general solution is y = c 1 e x + c e x + c sin x + c cos x. 1

2 e) y 6) + 8y ) + 16y = 0 Solution: This DE has the characteristic equation r 6 + 8r + 16r = 0, which factors as r r + ) = 0. Then r = 0 is a double root, and the pair r = ±i is also a double root, so the general solution is y = c 1 + c x + c + c x) cos x + c 5 + c 6 x) sin x.. Find solutions to the following IVPs: a) y y + y = 0, y0) = 1, y 0) = 0 Solution: The DE has the characteristic equation r r + = 0, and so has roots r = 1 and r =. Hence, its general solution is y = c 1 e x + c e x. We match this solution and its derivative, y = c 1 e x + c e x, to the initial conditions y0) = 1, y 0) = 0 at x = 0. We then obtain the linear system c 1 + c = 1, c 1 + c = 0. Then c 1 = c ; plugging this into the first equation, c = 1, so c = 1 and c 1 =. Therefore, the solution to the IVP is y = e x e x. b) 9y ) + 1y + y = 0, y0) = 0, y 0) = 1, y 0) = 10 Solution: The DE has the characteristic equation 9r + 1r + r = 0, which factors as rr + ) = 0. Hence, r = 0 is a single root, and r = / is a double root, so the general solution is y = c 1 + c e x/ + c xe x/. Taking derivatives, y = c e x/ + c 1 x ) e x/, y = 9 c e x/ + c + 9 x ) e x/. Evaluating these functions at x = 0 and matching the initial conditions, we obtain the linear system c 1 + c = 0, c + c = 1, 9 c c = 10. Then c = 1 + c, so from the third equation, 9 c 8 9 c = 10, and c = 1 9 = 1. Then c = 1 1 = 6, so c 1 = c = 1. Hence, the solution to the IVP is y = 1 1 e x/ 6xe x/.

3 . For each of the DEs below, find a particular solution to it: a) y 6y + 8y = x + 5 Solution: From Problem 1a), we see that the roots of the associated homogeneous DE are r = and r =, neither of which corresponds to the polynomial function x + 5. Therefore, we do not need to modify the guess y = Ax + B to avoid colliding with terms in the complementary solution. Plugging this guess for y into the nonhomogeneous DE, we have 0 6A) + 8Ax + B) = x + 5, so equating the constant terms and the coefficients on the x terms, we have that 8A = and 8B 6A = 5. Then A = 1/, so 8B = 5 + = 8, and B = 1. Hence, a particular solution is y = 1 x + 1. b) y ) + y = 1x 16 8e x Solution: Writing the characteristic equation for the associated homogeneous DE, we have r + r = r r + ) = 0. This equation has a double root at r = 0 and a pair of complex roots r = ±i, so the complementary solution is y c = c 1 + c x + c cos x + c sin x. From f x) = 1x 16 8e x, we might initially guess a particular solution of the form y = Ax + B + Ce x, but the first two terms overlap with terms in y c. Hence, we multiply only those terms by x to avoid the overlap, obtaining y = Ax + Bx + Ce x as the form of a solution. Then y = 6Ax + B + Ce x and y ) = 16Ce x, so the DE becomes 16Ce x + Ax + 8B + 16Ce x = 1x 16 8e x. Then C = 8, so C = 1, A = 1, so A = 1, and 8B = 16, so B =. Consequently, a particular solution is y = 1 x x 1 ex. c) y + y y = xe x Solution: The roots of the characteristic equation for this DE are r = and r = 1, so the complementary solution is y c = c 1 e x + c e x. From the xe x in the f x) function, we might guess y = Axe x + Be x as a particular solution, but the lower-order term overlaps with the complementary solution. Hence, we multiply these terms by x to separate them from y c, and the form of our guess is y = Ax e x + Bxe x.

4 Then its derivatives are y = Ax x )e x + B1 x)e x, y = A9x 1x + )e x + B9x 6)e x. Plugging these into the original nonhomogeneous DE, and observing that the x e x terms all cancel, we have A 1x + )e x + B9x 6)e x + Axe x + B 6x)e x Bxe x = xe x. Grouping the coefficients on the xe x and e x terms separately, we have the linear system 8A =, A B = 0. Then A = 1/, so B = 1 A = 1, and a particular solution is y = 1 x e x + 1 xe x.. Consider a mass of kg attached to a spring with spring constant 18 N/m. Find the displacement xt) with the initial conditions x0) = m, x 0) = 9 m/s, assuming the following damping c is present. If the system exhibits periodic behavior, find its possibly time-varying) amplitude and period. a) c = 0 Solution: Since c = 0, there is no damping is the system. We compute the circular frequency ω 0 of the system to be k/m = 18/ = rad/s, so the general solution for the displacement is of the form xt) = A cos t + B sin t. Its velocity is then x t) = A sin t + B cos t. Matching these functions to the initial conditions x0) = and x 0) = 9 at t = 0, we have A =, B = 9, so A = and B =, and therefore xt) = cos t + sin t. The amplitude is given by C = A + B = = 5, and the period T = π/ω 0 = π/ s. b) c = Solution: We now have damping with c =. The characteristic equation for the DE mx + cx + kx = 0 is r + r + 18 = 0, so we check the discriminant of this quadratic polynomial to determine whether we obtain real or complex roots. Since this is c km = 16 )18) = 16 1 = 18, which is negative, we will get complex roots, with values r = ± 18 ) = ± 8 = 1 ±.

5 Therefore, the general solution for the displacement and its velocity is xt) = Ae t cos t + Be t sin t x t) = Ae t cos t sin t) + Be t sin t + cos t), which we match to the initial conditions x0) = and x 0) = 9 to obtain the system A =, A + B = 9. Then B = 9 + = 1, so B = 1/, and the displacement is xt) = e t cos t + 1 e t sin t. We have that C = A + B = /8 = 97/8 = of this solution is π/ω 1 = π/., so the amplitude e t. Since the pseudofrequency is ω 1 =, the period is c) c = 1 Solution: We check the discriminant c km again when c = 1: now it is 1 1 = 0, so we expect a double root at r = c/m = 1/ =. Hence, the general forms of the displacement and the velocity are xt) = c 1 e t + c te t, x t) = c 1 e t + c 1 t)e t. Matching these to the initial conditions x0) = and x 0) = 9, we have the system c 1 = and c 1 + c = 9, so c = = 1. Thus, the displacement is xt) = + 1t)e t. There is no trigonometric component to the solution, so there is no amplitude or period. d) c = 0 Solution: For c = 0, we now expect an overdamped system with real roots given by r = 0 ± 0 1 = 0 ± 56 0 ± 16 = 5 ±, so the roots are r = 1 and r = 9. Therefore, the general forms of the displacement and the velocity are xt) = c 1 e t + c e 9t, x t) = c 1 e t 9c e 9t. Matching these to the initial conditions x0) = and x 0) = 9, we have the system c 1 + c =, c 1 9c = 9. Adding these equations, 8c = 1, so c = 1/8, and c 1 = c = 5/8. Thus, the displacement is xt) = 5 8 e t 1 8 e 9t, and as in the critically damped case there is no amplitude or period. 5

6 5. Find a differential equation with the general solution y = c 1 + c x)e x + c e x cos x) + c e x sin x). Solution: From the form of this general solution, we expect the characteristic equation of the DE to have a double root at r = and a pair of complex roots r = ± i. Then the polynomial in the characteristic equation is r ) r + i)r + + i) = r 6r + 9)r + r + 6) = r )r )r + 6 6)r + 5 = r r 9r + 5. Thus, the differential equation y ) y ) 9y + 5y = 0 or any constant multiple of it) corresponds to this characteristic equation, and hence has this general solution. 6. a) Show that the functions y 1 = e x and y = sin x are linearly independent. Solution: We compute the Wronskian Wx) of these two functions: Wx) = y 1 y y 1 y = e x sin x e x cos x = e x cos x + e x sin x = e x cos x + sin x). Since Wx) is not identically 0 on the real line as, for example, W0) = e ) = 1), these functions are linearly independent. b) Show that the functions y 1 = 1 + tan x, y = tan x, and y = sec x are not linearly independent. Solution: The easy way to show these function are linearly independent is to note that 1 + tan x = sec x, so y 1 = y. Thus, y 1 + y is a nontrivial linear combination of these functions equal to 0. If we do not notice this fact, we proceed by computing the derivatives of y 1, y, and y, using that tan x) = sec x and sec x) = sec x tan x: y 1 = sec x tan x, y 1 x tan x) + sec x) = sec x tan x + sec x y = sec x tan x, y = 8 sec x tan x sec x y = sec x tan x, y = sec x tan x + sec x The Wronskian of these solutions is then the determinant y 1 y y Wx) = y 1 y y y 1 y y 1 + tan x tan x sec x = sec x tan x sec x tan x sec x tan x sec x tan x + sec x 8 sec x tan x sec x sec x tan x + sec x 6

7 We expand this determinant along the first row, but first we use the properties of the deteriminant to extract the common factors of the second and third rows: Wx) = sec x tan x) sec x tan x + sec 1 + tan x tan x sec x x) = 8 sec x tan x + sec 6 x tan x) 1 + tan ) 1 1 tan x) sec x 1 ) 1 = 8 sec x tan x + sec 6 x tan x) ) = 0. Since the Wronskian is identically 0, the functions are linearly dependent. 7. Find the form of a particular solution to the DE y ) y + y = 6e x + e x sin x x, but do not determine the values of the coefficients. Solution: We first determine the complementary solution y c, the general solution of the associated homogeneous equation. The characteristic equation is r r + r = rr r + ) = 0. Thus r = 0 is a root, as are r = ± 8 = 1 ± i, so y c = c 1 + c e x cos x + c e x sin x. Examining the form of the forcing function f x), we would expect a particular solution of the form Ae x + Be x cos x + Ce x sin x + D + Ex + Fx, but the Be x cos x, Ce x sin x, and D terms overlap with y c. Therefore, we multiply these terms and the ones related to them by enough powers of x to prevent the overlap: y p = Ae x + Bxe x cos x + Cxe x sin x + Dx + Ex + Fx. 8. Consider the nonhomogeneous DE x y + xy + x 1 )y = 8x/ sin x. a) Verify that y 1 = x 1/ cos x and y = x 1/ sin x are linearly independent solutions to the associated homogeneous DE. Solution: We compute the first and second derivatives of these solutions: y 1 = 1 x / cos x x 1/ sin x y 1 = x 5/ cos x + x / sin x x 1/ cos x y = 1 x / sin x + x 1/ cos x y = x 5/ sin x x / cos x x 1/ sin x 7

8 We plug these into the associated homogeneous DE x y + xy + x 1 )y = 0: x y 1 + xy 1 + x 1 ) y 1 ) = x x 5/ cos x + x / sin x x 1/ cos x + x 1 ) x / cos x x 1/ sin x + x ) 1 ) x 1/ cos x = 1 1 ) x 1/ cos x + 1 1) x 1/ sin x ) x / cos x = 0 x y + xy + x 1 ) y ) = x x 5/ sin x x / cos x x 1/ sin x = + x 1 ) x / sin x + x 1/ cos x + x ) 1 ) x 1/ sin x 1 1 ) x 1/ sin x ) x 1/ cos x ) x / sin x = 0 We also check their linear independence by computing the Wronskian: Wx) = x 1/ cos x x 1/ sin x 1 x / cos x x 1/ sin x 1 x / sin x + x 1/ cos x = 1 x sin x cos x + x 1 cos x + 1 x sin x cos x + x 1 sin x = 1 x Since this function is not identically 0, the functions are linearly independent. b) Use variation of parameters to find a general solution to the nonhomogeneous DE. Hint: Use the identities sin x cos x = 1 sin x and sin x = 1 1 cos x).) Solution: We use the formula of variation of parameters to determine a particular solution to the nonhomogeneous DE. First, we must normalize the DE, so that y has a coefficient of 1: y + 1 x y ) x y = 8x 1/ sin x. Then f x) = 8x 1/ sin x, so the formula for variation of parameters is y f x) y = y 1 Wx) dx + y y1 f x) Wx) dx x = x 1/ 1/ sin x)8x 1/ sin x) cos x dx 1/x x + x 1/ 1/ cos x)8x 1/ sin x) sin x dx 1/x 8

9 Simplifying and applying the identities sin x cos x = 1 sin x, sin x = 1 1 cos x), and cos x = cos x 1, y = x 1/ cos x 8 sin x dx + x 1/ sin x 8 sin x cos x dx = x 1/ cos x 1 cos x) dx + x 1/ sin x sin x dx = x 1/ cos xx sin x) + x 1/ sin x cos x) = x 1/ cos xx sin x cos x) + x 1/ sin x cos x 1)) = x 1/ cos x + x 1/ sin x. The second term is already part of the complementary solution, so we can reduce this particular solution to y = x 1/ cos x. 9. A spring is stretched 6 inches by a mass m that weighs 8 lb. The mass is attached to a dashpot that has a damping force of lb-s/ft, and an external force of cos t lb acts on it. a) Describe the steady state response of the system that is, the particular solution to the nonhomogeneous equation). Solution: We compute the parameters associated to the system: the mass is m = 8/ = 1/ slug, assuming g = ft/s, and the spring constant is k = 8/0.5) = 16 lb/ft. The damping constant is c =, as given above. On the forcing side, the amplitude is F 0 =, and the frequency is ω =. Since the system is damped, the pure-sinusoidal forcing function will not overlap with the transient solution, so we expect the steady-state function will be of the form xt) = A cos t + B sin t. Plugging this and its derivatives into the nonhomogeneous DE mx + cx + kx = cos t, we have Ak m) + Bc) cos t + k m)b Ac) sin t = cos t. Then Ak m) + Bc = and k m)b Ac = 0, which we solve to obtain A = k m) k m) + c, B = 8c k m) + c. Plugging in k = 16, m = 1, and c =, k m = 15 and c =, k m) + c = = 1, so A = 60/1 and B = 16/1. Then the steady state solution is xt) = cos t + sin t

10 We may also unify this into a single oscillation C cost α), with Then C = C = A + B = k m) + c, tan α = B A = c k m. =, α = tan , xt) = cost α). 1 b) Find the value of the mass m that maximizes the amplitude of this response, with all other parameters remaining constant. What is this maximum amplitude? What does this mass weigh in pounds? Solution: We see that, as a function of m, the amplitude C is maximized when the denominator, or its square k m) + c, is minimized. Since this is a sum of two squares, only one of which contains m, this is minimized when that term is 0. That occurs when k m = 0, so m = k/ = 16/ = slug. This mass weighs 18 lb, and the corresponding amplitude is C = /cω = / = 1 ft. 10