SMN 3043 ORDINARY DIFFERENTIAL DIFFERENTIAL EQUA EQUATIONS ASSIGNMENT 3
BUNGEE JUMPING NAME
MATRIC MA TRIC NO.
NOORUL ASMA BINTI ABDUL SAMAT SAMAT
D20092036907 D2009203 6907
NUR HAFIZAH IZZATI IZZATI BINTI MD RASIP
D20111048854
NORFAEZAH NORFAEZAH BINTI HAMZAH
D20111048860
NOR ATIQAH FATIHAH BINTI ABDULLAH ABDULL AH
D20111048872 D2011104 8872
1.0 TWO METHODS TO SOLVE SOLVE HIGHER HIGHER ORDER ODE 1) Variat ariatio ion n of Par Parame amete terrs 2) Eul Euler-C er-Cau auch chy y ODE ODE
1.0 TWO METHODS TO SOLVE SOLVE HIGHER HIGHER ORDER ODE 1) Variat ariatio ion n of Par Parame amete terrs 2) Eul Euler-C er-Cau auch chy y ODE ODE
Variation of Parameters Consider the differential differential equation, Assume that y 1(t) and y 2(t) are a fundamental set of solutions for Then a particular solution to the nonhomogeneous differential differential equation is,
Example 1 •
Find a general solution to the following differential equation
2 y 18 y 6 tan(3t ) •
The differential equation the actually be solving is y 9 y 3 tan(3t )
Cont… •
Complementary solution is: yc (t ) c1 cos(3t ) c2 sin(3t )
•
So, we have
y1 (t ) cos(3t ) y2 sin(3t )
•
The Wronskian of these two function is W
cos(3t )
sin(3t )
3sin(3t ) 3cos(3t )
3cos2 (3t ) 3sin 2 (3t ) 3
Cont…. •
Particular solution: y p (t ) cos(3t )
cos(3t ) cos(3t )
2 sin (3t )
cos(3t )
3sin(3t ) tan(3t ) 3
dt sin(3t )
3cos(3t ) tan(3t ) 3
dt
dt sin(3t ) sin(3t )dt
1 cos 2 (3t ) cos(3t )
dt sin(3t ) sin(3t )dt
cos(3t ) sec(3t ) cos(3t )dt sin(3t ) sin(3t )dt
cos(3t ) 3 cos(3t ) 3
ln sec(3t ) tan(3t ) sin(3t ) ln sec(3t ) tan(3t )
sin(3t ) 3
cos(3t )
Cont…. •
General solution
y(t ) c1 cos(3t ) c2 sin(3t )
cos(3t ) 3
ln sec(3t ) tan(3t )
Example 2 Find the general solution to 2 ty (t 1) y y t
Given that:
y1 (t ) e
t
y2 (t ) t 1 Form a fundamental set of solutions for the homogeneous differential equation
Solution •
First, need to divide out by a t. 1 1 y 1 y y t t t
•
The Wronskian for the fundamental set of solution is W
et
t 1
et
1
t t t e e ( t 1) te
Cont… •
The particular solution is: t ( t 1) t e (t ) t Y p (t ) e dt (t 1) dt t t te te
et (t 1)et dt (t 1) dt t
e (e (t 2)) (t 1)t t
t 2 2t 2 •
General solution is: y(t ) c1et c2 (t 1) t 2 2t 2
Cauchy-Euler •
The Cauchy-Euler equation has the form n (n) n 1 ( n 1) ...... an1 xy an y b( x) a0 x y a1 x y x e
•
Where
•
k ( k ) x y Each term contains
•
t x e The transformation reduces the equation
t
are constants.
to a linear ODE with constant coefficient in the variable t. Notice that we assume x>0, and t=ln x. •
Using chain rule, since y function of t through x.
Example Find the general solution 3 x y 2 xy 2 y x , x 0 2
Solution: •
Taking the transformation x et the equation reduces to: yt 3 yt 2 y e3t
Solution •
•
Corresponding homogeneous equation: yt 3 yt 2 y 0 Characteristic equation: 2 r 3r 2 (r 2)(r 1) 0
r1 2, r 2 1 •
Fundamental set of solution: F e , e 2t
t
Cont…. •
•
•
•
•
Complementary solution: 2t t yc (t ) c1e c2e Non-homogeneous term is b(t ) e
3t
The UC set of
3t
e is
S1 e3t
The candidate for particular solution is: 3t y p (t ) Ae Computing the derivatives yp( x) 3 Ae3t yp( x) 9 Ae
3t
Cont… •
Substituting into the equation 9 Ae 9 Ae 2 Ae e 3t
A
3t
3t
1 2
y p (t )
1 2
e
3t
y (t ) c1e c2e 2t
•
t
General solution: y( x) c1 x c2 x 2
1 2
1 2
e
3t
x3
3t
2.0 Application of Higher Order Differential Equations Electric Circuit Q = Charge Current, I =
L = Inductance R = Resistance
= Elastance
= Electromotive Force
In the figure 1, it contains an electromotive force E (supplied by battery or generator) a resister, an inductor, L , and a capacitor in series C . If the charge on the capacitor at time, t is = the current is the root of change of Q with respect to : =
.
Kirchhoff’s voltage law says that the sum of these voltage drops is equal to supplied voltage:
= () Since =
, equation becomes
= ()
Equation 1
The charge, and current, are known at time 0, then we have initial conditions,
0 = therefore, 0 = = Initial value problem can be solved by methods of Additional: Non homogeneous linear equations. A differential equation for the current can be obtained by differentiating Equation 1 with respecting to and remembering that =
= ()
Question: Find the charge and current at time in the circuit of figure if = 40Ω, = 1, = 16 × 10− , = 100 cos 10 and the initial charge and current are both 0. From Equation 1 with given values of ,, and (),
40
625 = 100 cos
Auxiliary question is 40 625 = 0 with root,
=
−± −
= 20 ± 15
Solution of complementary equation is
= − ( cos 15 sin 15) For the method of undetermined coefficients we try particular solution,
= cos 10 sin 10
= 10 sin 10 10 cos 10 = 100 cos 10 100 sin 10
Substituting into Equation 2, we have
100 cos 10 100 sin 10 40(10 sin 10 10 cos 10) 625( cos 10 sin 10) = 100 cos 10 or
(525 400) cos 10 400 525 sin 10 = 100 cos 10 Equating coefficient, we have
525 400 = 100 and 4000 525 = 0 Or
21 16 = 4 and 16 21 = 0
Solution of this system is,
=
and =
.
So, particular solution is
=
(84 cos 10 64 sin 10)
General solution is
= = − cos 15 sin 15 (21 cos 10 16 sin 10)
Imposing the initial condition 0 = 0, we get
0 =
= 0, =
To impose other initial condition, we differentiate to find the current
= = − 20 15 cos 15 15 20 sin 15 40 21 sin 10 16 cos 10 697 0 = 20 15
= 0, =
∴ Formula of charge is =
[ (63 cos 15
(21
10
116 sin 15)
16 i 10)]
Expression for current is
1 [ − 1920 cos 15 13060 sin 15 = 2091 120 21 sin 10 16 cos 10 ]∎
PROBLEM 1 QUESTION: Solve the equation = for , given that you step off the bridge-that is no jumping, no diving! “ Stepping off” means that the initial conditions are 0 = 100, 0 = 0. Use = 160, = 1, and = 32.
SOLUTION •
•
We apply the theorem from subtopic 4.3 Homogenous Linear Equation With Constant Coefficients (Second-Order) by let as any particular solution on an interval I. To solve a non-homogenous differential equation, we need to find:
1) that is a fundamental set of solutions that form when and are linearly independent. 2) then, we find the .
1) •
Since it given the value of and , so the value of are:
= 160 , 160 = 160 = 32 =5
= 32
= 5 = 160 5 = 0 5 1 = 0 =0 ,
(a)
5 1 = 0 5 = 1 =
=
−
= 160 =
= 0 = 0 •
When substitute into (a);
5 0 0 = 160 0 0 = 160
not supposed to get this answer
•
Let = , as variable,
= = 0 5 0 = 160 = 160 = 160 =
−
160
(b)
2) Let = Given 0 = 100 •
−
160 = 100
−()
160 = 100 = 100
Given 0 = 0
=
−
−
160
= 160 1 − 160 = 0 5 = 160 = 800
(1)
•
•
After get the value of , substitute the value into (1)
= 100 800 = 100 = 900 Then substitute the value of and into (b) () =
−
160
() = 900 800 Let =
() = 900 800
−
−
160
160
PROBLEM 2 QUESTION: Use the solution from Problem 1 to compute the length of time you free-fall (that is, the time it takes to go to the natural length of the cord:100 feet)
SOLUTION •
From Problem 1,
= 100 = 900 − 800
800
−
−
− 800
160 = 100
160 = 100 900 160 = 1000
−
=
=
−
•
5 1 = 4 5
ln both side to eliminate exponent; − ln
5 1 = ln( ) 4 5 =
ln( )
= ln 25 ln(4)
1 ln(4) = ln 25 5 1 ( ln4) = ln 25 5
=
− +
= 2.71
= 2.71s
PROBLEM 3 QUESTION: Compute the derivative of the solution you found in Problem 1 and evaluate it at the time you found in Problem 2. You have found your downward speed when you pass the point where the cords starts to pull.
SOLUTION •
From the answer in Problem 1,
() = 900
− 800
160
We need to find it derivatives;
() = 900 =
160
− (800 ) 1 6 0
= 160 = 160
− 800
= 160(
160
−
160
−
−
1)
•
Substitute the value of found in Problem 2 into the derivatives that already calculate before;
= 2.71s
− = 160( 1) − (.) = 160(
2.71 2.71 = 66.9465
2.71 = 66.9465 ft/s
1)
PROBLEM 4 Solve the initial-value problem
= , = 0, = For now you may use the value = 14 but eventually you will need to replace this number with the values of for the cords you brought. The solution represents your position below the natural length of the cord after it starts to pull back.
SOLUTION Let = 2.71 and = 66.95 Have, = 160, = 1 and = 32, = 14
= 1 1 14 =1 32 160 160 160 14 = 160 32 5 14 = 160
Solve for Auxillary Equation: 5 14 = 0
= = =
−± −()() () −± − −±
Compare with =∝ ± We have, ∝=
∴ =
−
and
cos
=
sin
Solve for ,
= = 0 = 0 5 0 0 14 = 160 =
=
∴ =
The solution for is = So, =
−
cos
sin
We know, = 0 and = 2.71
2.71 =
. −
279(2.71) 279(2.71) 80 cos sin =0 10 10 7
0.76262 0.18473 0.98279 80 0.14088 0.74950 = 7 = 81.12274 5.32013
80 =0 7
And we have, = , = 2.71 and =66.95
279 279 279 279 = sin cos 10 10 10 10 1 − 279 279 cos sin 10 10 10 −
2.71 279 279 × 2.71 279 279 × 2.71 = sin cos 10 10 10 10 1 − 279 × 2.71 279 × 2.71 cos sin = 66.95 10 10 10 . −
1.26599 0.16036 = 66.95
Substitute value into the equation
1.26599(81.12274 5.32013 ) 0.16036 = 66.95 = 5.43746 And substitute back into , we get,
= 52.19475 Hence,
=
−
279 279 80 52.19475 cos 5.43746 sin ∎ 10 10 7
PROBLEM 5 Compute the derivative of the expression you found in Problem 4 and solve for the value of t where the derivative is zero. Denote this time as t 2. Be careful that the time you compute is greater than - there are several times when your motion stops at the top and bottom of your bounces ! After you find t 2, substitute it back into the solution you found in Problem 4 to find your lowest position
SOLUTION =
−
=
−
80 cos sin 7 sin cos
1 − cos sin 10 We have, = 0
−
−
sin cos
1 − cos sin = 0 10
sin cos
1 − = cos sin 10
sin cos
1 = cos sin 10
1 1 sin cos = 0 10 10 Substitute values , and
87. 87.72 7261 617 7 sin sin 3.86 3.862 287 co coss = 0 87. 87.72 7261 617 7 sin sin = 3.86 3.862 287 co coss sin = 0.04403 cos
tan = 0.04 0.0440 403 3 = tan− 0.04403 = 0.04400 Substitute value of
= 0.02634 ∎
Find the lowest position,
=
−
80 cos sin 7
Substitute value = 0.02634 So, 0.02 0.0263 634 4 = 63.6 63.6741 7417f 7ftt ∎
PROBLEM 6 (CAS) : You have brought a soft bungee cord with k = 8.5, a stiffer cord with k = 10.7, and a climbing rope for which k = 16.4. Which, if any, of these cords can you use safely under the given conditions?
Solution •
Problem 6 has been marked as computer problem. Therefore, we use the tool to plot solutions for the given k values (k – spring constant).
•
•
•
•
•
The horizontal line at the top represents the water.
Note that, the coordinate system is inverted – the positive direction is downwards. And remember that the jumper is 6 feet tall.
The value of k only can be set in positive value only. Click the Play button to have the bungee jumper “step-off” the bridge at x(0) = -100
•
•
Given soft bungee cord, k = 8.5 (set k = 8.6) and from the problem 1, we know that w = 160 Figure shows the position of the jumper.
•
•
•
From the figure, we can see that the river is 250 feet below and x (0) = -100. When we click the Play button, it shows that the jumper does not get wet. Then, from the graph, it shows that value of x is about 68
Given stiffer cord, k = 10.7 ( k = 10.8) and from the problem 1, we know that w = 160. Figure shows the position of the jumper. The animation of the tool shows that the jumper also does not get wet. Then, from the graph, it shows that value of x is about 58 •
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•
•
•
Given climbing rope, k = 16.4 and from the problem 1, we know that w = 160.
•
Figure shows the position of the jumper.
•
For this climbing
rope, it shows that the jumper also
does not get wet. •
And from the
graph, x is about
48.
•
•
•
•
From above, the value of k for climbing rope is the largest among the three cord that is k = 16.4. Followed with stiffer cord that is k = 10.7, while soft bungee jumping is the smallest that is k = 8.5. But we also need to consider the type of cord to avoid unpleasantness associated with an unexpected water landing. The weakness of using stiffer cord is the cord is too stiff, then your body will no longer form a connected set after you hit the end of the cord. While, The weakness of using climbing rope for the bungee jumping is the rope does not have the spring so it might cause the rope to broke.
•
•
•
Therefore, the soft bungee cord is the suitable cord to use for the bungee jumping. It is because, when the person jumps, the cord stretches and the jumper flies upwards again as the cord recoils, and continues to oscillate up and down until all the energy is dissipated. Hence, we can conclude that, the position of the jumper that indicate the relevant path as the jumper approaches the water is the one with k = 8.5 that is the soft bungee cord.
PROBLEM 7 You have a bungee cord for which you not determined the spring constant k. To do so, you suspend a weight of 10 pounds from the end of the 100-foot cord, causing it to stretch 1.2 feet. What is the value of k for this cord?
SOLUTION
Given weight, mg = 10 and x(t) = 1.2 •
We can use the equation of net force : ′
•
We know that, b(x) = -kx for x> 0 mx’’ = mg – kx - ′
Since the question ask the value of k, then rewrite the equation into : ′
Then, x(t) = 1.2, x’(t) = 0, x”(t)= 0.
Substitute all the value in the equation, mx” = mg – kx -
kx = mg - - mx” k(1.2) = 10 – 0 – 0 k=
.
k = 8.3 #
Therefore, the value of k for this cord is 8.3
Problem 8 (CAS) What would happen if your 220-pound friend uses the bungee cord whose spring constant is k = 10.7
•
•
By using the tool. Set the weight , mg= 220 and k = 10.7 (k = 10.8). Click the Play button to “step-off” . We can see that our friend is at least going to get wet.
•
•
•
•
It is because, before the jumping, the velocity, x’ = 0. After the jumping, the velocity change and the cord stretch into x = 68. Remember that our friend is about 6 feet tall. Therefore, we can conclude that at x = 68 it will cause our friend to become wet.
Problem 9 If your heavy friend wants to jump anyway, then how short should you make the cord so that he does not get wet?
By using the tool. Set the weight of our friend into the highest weight in the too that is 250 and set first we set k = 10.7 (k = 10.8) Click the Play button to “step-off” . We can see that our friend is going to get wet. It is because the bigger the weight of the jumper, the bigger the stretch of the cord. Hence, when the end of the cord scrapes the water, our friend is going to get wet. •
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•
•
•
•
•
•
•
To overcome him to not become wet, we can work backwords from the ending point to the initial condition. When the cord is shortened to about 96 feet (means that the initial condition is changed to x(0) = -96), then our friend seems to be able to jump without getting wet. Other than that, we can also change the value of k. Change the value of k become larger and the value of k is about 14. (k = 14). Hence, we can see that our friend does not get wet.
Reference Book •
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Curtis F. Gerald, Patrick O. Wheatley. (2004). Applied Numerical Analysis (Seventh ed.). USA: Greg Tobin. Richard L. Burden, J. Douglas Faires. (2005). Numerical Analysis (8th ed.). USA: Bob Pirtle. Zill, D. G. (2009). A First Course in Differential Equations With Modelling Applications (Ninth ed ed.). USA: Brooks/Cole Cengage Learning.
