Math 38 Exam II Practice Problems This review should not be used as your sole source for preparation for the exam. You should also re-work all examples given in lecture and all suggested homework problems.. Find the general solution of. Find the general solution of y + y y = 9e t. y + 9y = 3 sin(3t). 3. Find the general solution of y + 4y 5y = 8e 5t. 4. Find the general solution of y + 4y = tan(t), < t < π/4. 5. A mass of 3 kg stretches a spring 6.5 cm. If the mass is set in motion from.5 m above its equilibrium position with a downward velocity of m/s, and if there is no damping, determine the position of the mass at any time. Find the frequency, period, amplitude, and phase of the motion. When does the mass first return to its equilibrium position? 6. A mass weighing 4 lb stretches a spring 3 in. The mass is attached to a viscous damper with a damping coefficient of lb s/ft. If the mass is set in motion from 9 in above its equilibrium position with an upward velocity of ft/s, determine the position of the mass at any time. Find the quasifrequency and quasiperiod. 7. A /5 kg mass stretches a spring 9.8 m. The mass is acted on by an external force of cos t N. If the mass is set in motion from its equilibrium position with a downward velocity of m/s, find the position of the mass at any time. Identify the transient and steady state solutions. Does the motion exhibit resonance or a beat? 8. A mass weighing 8 lb stretches a spring.8 feet. The mass is attached to a dashpot mechanism that has a damping coefficient of lb s/ft and is acted on by an external force of cos(t) lb. If the mass is set in motion from its equilibrium position with no initial velocity, find the position of the mass at any time. What is the resonance frequency for the system? 9. Use the definition of the Laplace transform to find L{sinh(at)}.. Use the definition of the Laplace transform to find L{f(t)}, where { 3, t <, f(t) = 6 t, t.
. Find the inverse Laplace transform of each function. 7 (a) F (s) = (s + 3) 3 s + (b) F (s) = s 4s + 3 4s + 3s + 9 (c) F (s) = (s )(s + 4s + 3) (d) F (s) = e s (4s + ) (s )(s + ) s (e) F (s) = (s + ). Sketch the graph of f(t) = { t, t <,, t. Express f(t) in terms of the unit step function u c (t). 3. Sketch the graph of f(t) = { 4, t < 3, t, t 3. Express f(t) in terms of the unit step function u c (t) and find its Laplace transform. 4. Use Laplace transforms to solve the initial value problem 5. Solve the initial value problem y y + y = cos t, y() =, y () =. y y = g(t), y() =, y () =, where g(t) = { sin t, t < π,, t π. 6. Solve the initial value problem y + 4y = δ(t 4π), y() = /, y () =. 7. Solve the initial value problem y + 4y + 5y = δ(t π/6) sin t, y() =, y () =. 8. Solve the initial value problem y + y = g(t), y() = 3, y () =. Express your answer in terms of a convolution integral.
Solutions Solutions may contain errors or typos. If you find an error or typo, please notify me at glahodny@math.tamu.edu.. Find the general solution of y + y y = 9e t. Consider the corresponding homogeneous equation y + y y =. If y = e rt, then we obtain the characteristic equation r + r =. The roots of the characteristic equation are Therefore, the homogeneous solution is Consider a particular solution of the form To find A, we calculate r = ( ± 4 + 4) = ±. y h (t) = c e ( + )t + c e ( )t. Y (t) = Ae t. Y (t) = Ae t, Y (t) = 4Ae t. Substituting into the differential equation, we obtain (4A + 4A A)e t = 7Ae t = 9e t. It follows that A = 9/7, and a particular solution is Y (t) = (9/7)e t. The general solution is y(t) = y h (t) + Y (t) = c e ( + )t + c e ( )t + 9 7 et. 3
. Find the general solution of y + 9y = 3 sin(3t). Consider the corresponding homogeneous equation y + 9y =. If y = e rt, then we obtain the characteristic equation r + 9r = r(r + 9) =. The roots of the characteristic equation are r = and r = 9. Therefore, the homogeneous solution is y h (t) = c + c e 9t. Consider a particular solution of the form To determine the coefficients, we calculate Y (t) = A sin(3t) + B cos(3t). Y (t) = 3A cos(3t) 3B sin(3t), Y (t) = 9A sin(3t) 9B cos(3t). Substituting into the differential equation and collecting terms, we obtain ( 9A 7B) sin(3t) + (7A 9B) cos(3t) = 3 sin(3t). Thus, A and B must satisfy the equations 9A 7B = 3, 7A 9B =. It follows that A = /3 and B = /. Therefore, a particular solution is The general solution is Y (t) = 3 sin(3t) cos(3t). y(t) = c + c e 9t 3 sin(3t) cos(3t). 4
3. Find the general solution of y + 4y 5y = 8e 5t. Consider the corresponding homogeneous equation y + 4y 5 =. If y = e rt, then we obtain the characteristic equation r + 4r 5 = (r + 5)(r ) =. The roots of the characteristic equation are r = 5 and r =. Therefore, the homogeneous solution is y h (t) = c e 5t + c e t. Since g(t) = 8e 5t and r = 5 is a single root of the characteristic equation, we consider a particular solution of the form To determine the coefficients, we calculate Y (t) = Ate 5t. Y (t) = (A 5At)e 5t, Y (t) = (5At A)e 5t. Substituting into the differential equation, we obtain 6Ae 5t = 8e 5t. It follows that A = 4/3, and a particular solution is Y (t) = (4/3)te 5t. The general solution is y(t) = c e 5t + c e t 4 3 te 5t. 5
4. Find the general solution of y + 4y = tan(t), < t < π/4. Consider the corresponding homogeneous equation y + 4y =. If y(t) = e rt, then we obtain the characteristic equation r + 4 =. The roots of the characteristic equation are r = ±i. Therefore, the homogeneous solution is y h (t) = c cos(t) + c sin(t). The Wronskian of y (t) = cos(t) and y (t) = sin(t) is W (y, y )(t) = cos(t) sin(t) sin(t) cos(t) =. Thus, y and y form a fundamental set of solutions of the homogeneous equation. Using the method of variation of parameters, let u (t) = sin(t) tan(t) dt = sin (t) cos(t) dt = (cos (t) ) dt cos(t) = (cos(t) sec(t)) dt = (sin(t) ln sec(t) + tan(t) ). 4 Similarly, let u (t) = cos(t) tan(t) dt = sin(t) dt = 4 cos(t). Thus, a particular solution is Y (t) = 4 cos(t)(sin(t) ln sec(t) + tan(t) ) sin(t) cos(t) 4 Y (t) = cos(t) ln sec(t) + tan(t). 4 Therefore, the general solution is y(t) = c cos(t) + c sin(t) cos(t) ln sec(t) + tan(t). 4 6
5. A mass of 3 kg stretches a spring 6.5 cm. If the mass is set in motion from.5 m above its equilibrium position with a downward velocity of m/s, and if there is no damping, determine the position of the mass at any time. Find the frequency, period, amplitude, and phase of the motion. When does the mass first return to its equilibrium position? Let us measure the displacement y in meters. Since there is no damping, the equation of motion is my + ky =. The stiffness k is determined from the statement that the mass stretches the spring by 6.5 cm, or.65 m. Thus, k = (3 kg)(9.8 m/s ).65 m = 48 N/m. Therefore, the equation of motion is 3y + 48y = y + 6y = The initial conditions are y() =.5 and y () =. If y = e rt, then we obtain the characteristic equation r + 6 =. The roots of the characteristic equation are r = ±4i. Thus, the general solution is y(t) = c cos(4t) + c sin(4t). Differentiating, we obtain y (t) = 4c sin(4t) + 4c cos(4t). Applying the initial conditions, we have y() = c =.5 y () = 4c = It follows that c = / and c = /. Therefore, the position of the mass is y(t) = cos(4t) + sin(4t). The frequency is ω = 4 rad/s, the period is T = π/ω = π/ s, the amplitude is R = / m, and tan δ = c /c =. Since c is positive and c is negative, δ is in Quadrant II, and the phase is δ = 3π/4 rad. 7
To determine when the mass will pass through the equilibrium position, y =, we first express y(t) in the form y(t) = cos(4t 3π/4). Then we must solve the equation This is satisfied when y(t) = cos(4t 3π/4) =. 4t 3π 4 = 4t = t = (n + )π (n + )π + 3π 4 (n + )π + 3π 8 6 where n is an integer. Substituting n = determines the first time t > when the mass crosses its equilibrium position: t = π/6 s. 8
6. A mass weighing 4 lb stretches a spring 3 in. The mass is attached to a viscous damper with a damping coefficient of lb s/ft. If the mass is set in motion from 9 in above its equilibrium position with an upward velocity of ft/s, determine the position of the mass at any time. Find the quasifrequency and quasiperiod. Let us measure the displacement y in feet. The equation of motion is To determine the mass m, we have my + by + ky =. m = w g = 4 lb 3 ft/s = lb s 8 ft. The stiffness k is determined from the statement that the mass stretches the spring 3 in, or /4 ft. Thus, k = 4 lb = 6 lb/ft. /4 ft Therefore, the equation of motion is 8 y + y + 6y = y + 6y + 8y = The initial conditions are y() = 3/4 and y () =. If y = e rt, then we obtain the characteristic equation r + 6r + 8 =. The roots of the characteristic equation are r = Thus, the general solution is Differentiating, we obtain ( 6 ± 56 5 ) = 8 ± 8i. y(t) = c e 8t cos(8t) + c e 8t sin(8t). y (t) = 8c e 8t cos(8t) 8c e 8t sin(8t) 8c e 8t sin(8t) + 8c e 8t cos(8t). Applying the initial conditions, we have y() = c = 3/4 y () = 8c + 8c = It follows that c = 3/4 and c =. Therefore, the position of the mass is y(t) = 3 4 e 8t cos(8t) e 8t sin(8t). The quasifrequency is β = 8 rad/s and the quasiperiod is T d = π/β = π/4 s. 9
7. A /5 kg mass stretches a spring 9.8 m. The mass is acted on by an external force of cos t N. If the mass is set in motion from its equilibrium position with a downward velocity of m/s, and there is no damping, find the position of the mass at any time. Identify the transient and steady state solutions. Does the system exhibit resonance or a beat? Let us measure the displacement y in meters. Since there is no damping, the equation of motion is my + ky = F (t). The stiffness k is determined from the statement that the mass stretches the spring 9.8 m. Thus, k = (/5 kg)(9.8 m/s ) = 9.8 m 5 N/m. Therefore, the equation of motion is 5 y + 5 y = cos t y + y = 5 cos t The initial conditions are y() = and y () =. The corresponding homogeneous equation is y + y =. If y = e rt, then we obtain the characteristic equation r + =. The roots of the characteristic equation are r = ±i. Thus, the homogeneous solution is y h (t) = c cos t + c sin t. Consider a particular solution of the form Y (t) = At sin t + Bt cos t. Then, Y (t) = A sin t + At cos t + B cos t Bt sin t, Y (t) = A cos t At sin t B sin t Bt cos t. Substituting into the differential equation and collecting like terms, we obtain Y + Y = 5 cos t A cos t B sin t = 5 cos t It follows that A = 5/ and B =. Therefore, a particular solution is Y (t) = 5 t sin t.
Thus, the general solution (and its derivative) are y(t) = c cos t + c sin t + 5 t sin t, y (t) = c sin t + c cos t + 5 sin t + 5 t cos t. Applying the initial conditions, we obtain Therefore, the solution is y() = c = y () = c = y(t) = sin t + 5 t sin t = ( +.5t) sin t. The transient solution is y h (t) = sin t and the steady state solution is Y (t) =.5t sin t. Since the frequency of the external forcing function is the same as the frequency of the transient solution, the system exhibits resonance. The graph of the solution is given below. 5 5 y -5 - -5 3 4 5 t
8. A mass weighing 8 lb stretches a spring.8 feet. The mass is attached to a dashpot mechanism that has a damping coefficient of lb s/ft and is acted on by an external force of cos(t) lb. If the mass is set in motion from its equilibrium position with no initial velocity, find the position of the mass at any time. What is the resonance frequency for the system? Let us measure the displacement y in feet. The equation of motion is To determine the mass m, we have my + by + ky = F (t). m = w g = 8 lb 3 ft/s = lb s 4 ft. The stiffness k is determined from the statement that the mass stretches the spring.8 ft. Thus, k = 8 lb = lb/ft..8 ft Therefore, the equation of motion is 4 y + y + y = cos(t) y + 4y + 4y = 8 cos(t) The initial conditions are y() = and y () =. The corresponding homogeneous equation is y + 4y + 4y =. If y = e rt, then we obtain the characteristic equation r + 4r + 4 =. The roots of the characteristic equation are r = Thus, the homogeneous solution is ( 4 ± 6 6 ) = ± 6i. y h (t) = c e t cos(6t) + c e t sin(6t). Consider a particular solution of the form Y (t) = A sin(t) + B cos(t). Then, Y (t) = A cos(t) B sin(t), Y (t) = 4A sin(t) 4B cos(t).
Substituting into the differential equation and collecting like terms, we obtain Y + 4Y + 4Y = 8 cos(t) (36A 8B) sin(t) + (8A + 36B) cos(t) = 8 cos(t) It follows that A = 4/85 and B = 8/85. Therefore, a particular solution is Y (t) = 4 85 sin(t) + 8 85 cos(t). Thus, the general solution (and its derivative) are y(t) = c e t cos(6t) + c e t sin(6t) + 4 8 sin(t) + 85 85 cos(t), y (t) = c e t cos(6t) 6c e t sin(6t) c e t sin(6t) + 6c e t cos(6t) + 8 36 cos(t) 85 85 sin(t). Applying the initial conditions, we obtain y() = c + 8 85 = y () = c + 6c + 8 85 = It follows that c = 8/85 and c = /55. Therefore, the solution is y(t) = 8 85 e t cos(6t) 55 e t sin(6t) + 4 8 sin(t) + 85 85 cos(t). The resonance frequency of the system is k γ r = m b m = 3 = 4. 3
9. Use the definition of the Laplace transform to find L{sinh(at)}. Recall that sinh(at) = (e at e at )/. The Laplace transform of f(t) = sinh(at) is defined as L{sinh(at)} = = e st sinh(at) dt e st (e at e at ) dt = (e (s a)t e (s+a)t ) dt = [ s a e (s a)t + ] s + a e (s+a)t = [ lim N s a e (s a)t + ] N s + a e (s+a)t = [ lim N s a e (s a)n + s + a e (s+a)n + s a ] s + a = ( s a ) s + a = ( ) a s a a = s a. In the above argument, the terms e (s a)n and e (s+a)n converge to zero as N if and only if s a > and s + a >, respectively. That is, if and only if s > a and s > a, or equivalently, if s > a. Therefore, L{sinh(at)} = a s a, s > a. 4
. Use the definition of the Laplace transform to find L{f(t)}, where { 3, t <, f(t) = 6 t, t. The Laplace transform of f(t) is defined as L{f(t)} = = e st f(t) dt 3e st dt + e st (6 t) dt = 3 ( s e st + s (t 6)e st + ) s e st = 3 ( s ( e s ) + lim N s (t 6)e st + ) N s e st = 3 ( s ( e s ) + lim N s (t 6)e sn + s e sn + 4 s e s ) s e s = 3 s ( e s ) + 4 s e s s e s = 3 s + s e s s e s, s >.. Find the inverse Laplace transform of each function. (a) F (s) = 7 (s + 3) 3 Using linearity of the inverse Laplace transform, we have { } L 7 = 7 { }! (s + 3) 3 L = 7 (s + 3) 3 t e 3t. 5
(b) F (s) = s + s 4s + 3 By completing the square of the quadratic in the denominator, we obtain F (s) = s + (s ) + 9 = (s ) + 5 (s ) + 9. Using linearity of the inverse Laplace transform, we have { } (s ) L {F (s)} = L + 53 { } 3 (s ) + 9 L (s ) + 9 = e t cos(3t) + 5 3 et sin(3t). (c) F (s) = 4s + 3s + 9 (s )(s + 4s + 3) Consider the partial fraction decomposition for F (s): 4s + 3s + 9 (s )(s + 4s + 3) = A s + Bs + C s + 4s + 3. Multiplying by the denominator, we obtain 4s + 3s + 9 = A(s + 4s + 3) + (Bs + C)(s ) Equating the coefficients, we have = (A + B)s + (4A B + C)s + (3A C). A + B = 4 4A B + C = 3 3A C = 9. It follows that A =, B =, and C = 7. Therefore, F (s) = s + s + 7 s + 4s + 3. By completing the square of the quadratic in the denominator, we obtain F (s) = s + s + 7 (s + ) + 9 = (s + ) + 3 + s (s + ) + 9. Using linearity of the inverse Laplace transform, we have L {F (s)} = { } { } { } L + L s + + L 3 s (s + ) + 9 (s + ) + 9 = e t + e t cos(3t) + e t sin(3t). 6
(d) F (s) = e s (4s + ) (s )(s + ) To use the translation property from Theorem 6.3., we first express F (s) as the product e s G(s), where G(s) = 4s + (s )(s + ). Consider the partial fraction decomposition for G(s): 4s + (s )(s + ) = A s + B s +. Multiplying by the denominator, we obtain Equating the coefficients, we have 4s + = A(s + ) + B(s ) = (A + B)s + (A B). A + B = 4 A B =. It follows that A = and B =. Therefore, G(s) = s + s +. Using linearity of the inverse Laplace transform, we have { } { } g(t) = L + L = e t + e t. s s + By Theorem 6.3., it follows that L {F (s)} = L {e s G(s)} = u (t)(e t + e (t ) ). 7
(e) F (s) = s (s + ) The function can be written as ( ) ( ) s F (s) = = G(s)H(s), s + s + where g(t) = cos t and h(t) = sin t. By the convolution theorem, we have f(t) = (g h)(t) = = t t cos(t τ) sin(τ) dτ [sin t sin(t τ)] dτ = t sin t [ 4 cos(t τ) ] t = t sin t 4 cos( t) + 4 cos(t) = t sin t. Here, we make use of the identity cos u sin v = [sin(u + v) sin(u v)]. (Note: On the exam, you may leave your answer in terms of the convolution integral). 8
. Sketch the graph of f(t) = { t, t <,, t. Express f(t) in terms of the unit step function u c (t). The graph of y = f(t) is shown below. 4 3.5 3.5 y.5.5 3 4 5 t The function f can be expressed as f(t) = t + ( t )u (t). 3. Sketch the graph of f(t) = { 4, t < 3, t, t 3. Express f(t) in terms of the unit step function u c (t) and find its Laplace transform. The graph of y = f(t) is shown below. 5 4 3 y 3 4 5 t 9
The function f can be expressed as To apply Theorem 6.3., we rewrite f as f(t) = 4 + (t 4)u 3 (t). f(t) = 4 + ((t 3) )u 3 (t) = 4 + (t 3)u 3 (t) u 3 (t). Using linearity of the Laplace transform and Theorem 6.3., we have L{f(t)} = 4L{} + L{u 3 (t)(t 3)} L{u 3 (t)} = 4 s + e 3s s e 3s s = (4 e 3s )s + e 3s s. 4. Use Laplace transforms to solve the initial value problem y y + y = cos t, y() =, y () =. Taking the Laplace transform of the differential equation, we obtain Using Corollary 6.., we have L{y } L{y } + L{y} = L{cos t}. s L{y} sy() y () [sl{y} y()] + L{y} = Applying the initial conditions and setting Y (s) = L{y}, we obtain (s s + )Y (s) s + = Y (s) = Y (s) = Expanding the right side using partial fractions, we have s s +. s s + s s s + + s (s + )(s s + ) s 3 s + s (s + )(s s + ). Y (s) = s 5 s + + 4s 6 5 s s + = s 5 s + + 4(s ) 5 (s ) +. Using linearity of the inverse Laplace transform, we obtain y(t) = L {Y (s)} = 5 (cos t sin t + 4et cos t e t sin t).
5. Solve the initial value problem y y = g(t), y() =, y () =, where g(t) = { sin t, t < π,, t π. To solve the problem, it is convenient to write g(t) = sin t u π (t) sin t = sin t + u π (t) sin(t π). Taking the Laplace transform of the differential equation, we obtain By Corollary 6.., we have L{y } L{y} = L{sin t} + L{u π (t) sin(t π)}. s L{y} sy() y () L{y} = s + + e πs s +. Applying the initial conditions and setting Y (s) = L{y}, we obtain Making use of partial fractions, we have (s )Y (s) = + e πs s + Y (s) = ( + e πs ) (s )(s + ). (s )(s + ) = 4(s + ) + 4(s ) (s + ). Using linearity of the inverse Laplace transform, we obtain { } L = (s )(s + ) 4 e t + 4 et sin t. By Theorem 6.3., the solution of this initial value problem is { } y(t) = L + e πs (s )(s + ) y(t) = 4 e t + 4 et sin t +u π (t) [ 4 e (t π) + 4 et π ] sin(t π).
6. Solve the initial value problem y + 4y = δ(t 4π), y() = /, y () =. Taking the Laplace transform of the differential equation, we obtain L{y } + 4L{y} = L{δ(t 4π)}. By Corollary 6.., we have s L{y} sy() y () + 4L{y} = e 4πs. Applying the initial conditions and setting Y (s) = L{y}, we obtain (s + 4)Y (s) s = e 4πs Y (s) = s (s + 4) + e 4πs s + 4. By Theorem 6.3., we have y(t) = { } { } s e L + L 4πs s + 4 s + 4 y(t) = cos(t) + u 4π(t) sin(t 8π).
7. Solve the initial value problem y + 4y + 5y = δ(t π/6) sin t, y() =, y () =. Taking the Laplace transform of the differential equation, we obtain By Corollary 6.., we have L{y } + 4L{y } + 5L{y} = L{δ(t π/6) sin t}. s L{y} sy() y () + 4[sL{y} y()] + 5L{y} = e (π/6)s sin(π/6). Applying the initial conditions and setting Y (s) = L{y}, we obtain (s + 4s + 5)Y (s) = e (π/6)s Y (s) = e (π/6)s Y (s) = e (π/6)s By Theorem 6.3., we have y(t) = { } e (π/6)s L (s + ) + (s + 4s + 5) (s + ) +. y(t) = u π/6(t)e (t π/6) sin(t π/6). 8. Solve the initial value problem y + y = g(t), y() = 3, y () =. Express your answer in terms of a convolution integral. Taking the Laplace transform of the differential equation, we obtain By Corollary 6.., we have L{y } + L{y} = L{g(t)}. s L{y} sy() y () + L{y} = G(s). Applying the initial conditions and setting Y (s) = L{y}, we obtain By the convolution theorem, we have (s + )Y (s) 3s + = G(s) Y (s) = 3s s + + G(s) s +. y(t) = 3 cos t sin t + t sin(t τ)g(τ) dτ. 3