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PRACTICE PROBLEMS CHAPTER 6 AND 7 I. Laplace Transform
- Find the Laplace transform of the following functions. (a) f^ t^ =sin^ ^2 t^ cos^ ^2 t^ (b) (^) f t =cos^2 3 t (c) (^) f t = t e^2 t^ sin 3 t (d) f^ t^ = t^ ^3 u 7 t^ (e) f t = t 2 u 3 t
(f) f^ t^ ={
1, if 0 ≤ t 2, t 2 − 4 t 4, if t ≥ 2
(g) f^ t^ ={
t , if 0 ≤ t 3, 5, if t ≥ 3 (h) f^ t^ = { 0, if t , t − , if ≤ t 2 0 , if t ≥ 2
(i) f^ t^ ={
cos t , if t 4, 0, if t ≥ 4 (j) f t =
t , if 0 ≤ t 1, e t , if t ≥ 1
- Find the inverse Laplace Transform: (a) F^ ^ s =^ 1 s 1 s 2 − 1 (b) F^ ^ s =^ 2 s 3 s 2 4 s 13 (c) F s = e −3s s − 2 (d) F s = 1 e − 2 s s 2 6
- The transform of the solution to a certain differential equation is given by X^ s =^ 1 − e − 2 s s 2 1 . Determine the solution x ( t ) of the differential equation.
- Suppose that the function y t satisfies the DE y ' ' − 2 y ' − y =1, with initial values, y 0 =−1, y ' 0 = 1. Find the Laplace transform of y t
- Consider the following IVP: y ' ' − 3 y ' − 10 y =1, y 0 =−1, y ' 0 = 2 (a) Find the Laplace transform of the solution y ( t ). (b) Find the solution y ( t ) by inverting the transform.
- Consider the following IVP: y^ '^ '^ ^4 y =^4 u 5 t^ ^ ,^ y^ ^0 =0,^ y^ '^ ^0 =^1 (a) Find the Laplace transform of the solution y ( t ). (b) Find the solution y ( t ) by inverting the transform.
- A mass m =1 is attached to a spring with constant k =5 and damping constant c = 2. At the instant t = the mass is struck with a hammer, providing an impulse p = 10. Also, x^ ^0 =^0 and x '(0)=0. a) Write the differential equation governing the motion of the mass. b) Find the Laplace transform of the solution x ( t ). c) Apply the inverse Laplace transform to find the solution. II. Linear systems
- Verify that x = e t
1
2 t e t
1
1 ^
is a solution of the system x^ '^ =
2 − 1
x e t
1
- Given the system (^) x ' = t x − y et^ z , y ' = 2 x t^2 y − z , z ' = e − t 3 t y t^3 z , define x , P( t ) and f t such that the system is represented as x ' =P t x f t
- Consider the second order initial value problem: u ' ' 2 u ' 2 u = 3 sin t , u 0 =2, u ' 0 =− 1 Change the IVP into a first-order initial value system and write the resulting system in matrix form.
- Are the vectors x 1 = 1 − 1 1 , x 2 = 0 1 1 and x 3 = 1 1 1 linearly independent?
5. Consider the system x ' =
− 2 − 6
x Two solutions of the system are x 1 = e t
− 2
1 ^
and x 2 = e − 2 t
1
(a) Use the Wronskian to verify that the two solutions are linearly independent. (b) Write the general solution of the system.
ANSWERS TO PRACTICE PROBLEMS CHAPTER 6 AND 7 I. Laplace Transform
- (a) Using the double angle trigonometric identity, the function f t can be rewritten as f t = 1 2 sin 4t . (^) Thus L { f t }= (^) s (^2) ^216 (b) Using the half angle trigonometric identity, the function f t can be rewritten as f t = 1 2 1 cos6t . (^) Thus L^ {^ f^ t^ }=^ 1
1 s s s 2
(c) Using the property L { t f t }=− F ' s with F^ ^ s =L^ { e 2 t sin 3 t }= 3 s − 2 2 9 yields L { t e 2 t sin 3 t }= 6 s − 2
s − 2
2
2 (d) f^ t^ =[ t −^7 ^10 ] u 7 t^ .^ Thus L^ {^ f^ t^ }= e − 7 s L { t 10 }= e − 7 s
1 s 2 ^ 10
s
(e) L^ {^ f^ t^ }= e − 3 s L { t 3 2 }= e − 3 s L { t 2 6t 9 }= e − 3 s
2 s 3 ^ 6 s 2 ^ 9
s
(f) f t = 1 u 2 t t 2
− 4 t 3 = 1 u 2 t [ t − 2
2
− 1 ]
Thus L^ {^ f^ t^ }=^ 1 s e − 2 s L { t 2 − 1 }= 1 s e − 2 s
2 s 3 −^ 1
s
(g) f^ t^ = t^ − u 3 t^ t^ −^5 = t − u 3 t^ [ t −^3 −^2 ].^ Thus L { f t }= 1 s 2 − e − 3 s L{ t − 2 }= 1 s 2 − e −3s
1 s 2 −^ 2
s
(h) f^ t^ = u t^ t −− u 2 t^ t −= u t^ t^ −− u 2 t^ t^ −^2 Thus L^ {^ f^ t^ }= e − s L{ t }− e − 2 s L { t }= e − s s 2 − e − 2 s
1 s 2 ^
s
(i) f^ t^ =cos^ t^ − u 4 t^ cos^ ^ t^ =cos t^ − u 4 t^ cos^ t −^4 ^ Thus L { f t }= s 2 s 2 − e − 4 s L{cos t }= s 2 s 2 − e − 4 s s 2 s 2 (j) f t = t u 1 t [ e t − t ]= t u 1 t [ e t − 1 1 − t − 1 − 1 ] Thus L { f t }= 1 s 2 e − s L { e t 1 − t − 1 }= 1 s 2 e − s
e s − 1 − 1 s 2 −^ 1
s
(a) Using PFD, F^ ^ s =−^ 1 4 1 s 1 − 1 2 1 s 1 2 ^ 1 4 1 s − 1
. (^) Thus f t =− 1 4 e − t − 1 2 t e − t 1 4 e t (b) F ( s ) can be rewritten as F^ ^ s =^ 2 s 3 s 2 2 9 = 2 s 2 − 1 s 2 2 9 = 2 s 2 s 2 2 9 − 1 3 3 s 2 2 9 .
Thus f^ t^ = e − 2 t
^2 cos^3 t^ −^
1 3
sin 3 t
(c) The inverse Laplace is u 3 t^ ^ f^ t −^3 ^ where f^ t^ =L − 1
1
s − 2 }
= e 2 t . Thus L − 1
e − 3 s
s − 2 }
= u 3 t e 2 t − 3 (d) F^ ^ s =^ 1
^6
s 2 6 e − 2 s
^6
s 2 6 thus L − 1 { F s }= 1
^6
sin 6 t
1
u 2 t sin 6 t − 2
- ^1 − u 2 t^ sin^ t
- Y^ s^ =^ − s 3 s 2 − 2 s − 1 1 s s 2 − 2 s − 1
- (a) Y^ s^ =^ 1 s s − 5 s 2 − 1 s 2 . (^) (b) y t =− 1 10 1 35 e 5 t − 13 14 e − 2 t
- (a) Y^ s^ =^ 1 s 2 4 e − 5 s
1 s − s s 2
. (^) (b) y t = 1 2 sin 2 t u 5 t [ 1 −cos 2 t − 10 ]
- (a) x ' ' 2 x ' 5 x = 10 t − (b) X^ s =^ 10 e − s s 2 2 s 5 = 5 e − s^2 s 1 2 4 (c) (^) x t = 5 u t e − t^ −^ sin 2 t −= 5 u t e − t^ ^ sin 2t II. Linear Systems
- Differentiating the given x yields x ' = e t
1
2 e t 2 t e t
1
3 e t 2 t e t 2 e t 2 t e
t
Substituting x into the right hand side of the DE yields:
2 − 1
e t 2 t e t 2 t e
t e
t
1
2 e t 4 t e t − 2 t e t 3 e t 6 t e t − 4 t e
t
e t − e
t =
3 e t 2 t e t 2 e t 2 t e
t = x^ '
- x =
x y
z ^
P t =
t − 1 e t 2 t 2 − 1 0 3 t t
f t =
0 0 e
− t
u'
v '
0 1
u
v
0
3 sin t ^
u 0
v 0
2
