# Bungee Jumping

Topics: Linear differential equation, Differential equation, Derivative Pages: 10 (2163 words) Published: May 12, 2013
1. Find two (2) methods (other than Undetermined Coefficients) that can be used to solve higher order ODE. For each method, solve one higher order ODE problem. Answer:
The general form of an ODE with order nis
Fx, yx, y''x,…,ynx=f(x)
As in the case of second-order ODEs, such an ODE can be classified as linear or nonlinear. The general form of a linear ODE of order is an(x)dnydxn+an-1(x)dn-1ydxn-1+…+a1(x)dydx+a0xy=f(x)
If f(x)is the zero function, the equation is said to behomogeneous. Many methods for solving higher order ODEs can be generalized to linear ODEs of ordern, where nis greater than 2. If the order of the ODE is not important, it is simply called a linear ODE. a). Variation of Parameters

Let’s,
an(x)dnydxn+an-1(x)dn-1ydxn-1+…+a1(x)dydx+a0xy=f(x)
Put it, in the standard form,
yn+Pxyn-1+…+Qxy'+Rxy=f(x)
Then, we assume that,
yP=u1y1+u2y2
By Cramer’s rule, the solution of the system,
y1u1'+y2u2'=0
y1'u1'+y2'u2'=f(x)
can be expressed in terms of determinants:
u1'=W1W= -y2 f(x)Wandu2'=W2W= -y1f(x)W
Where W=y1y2y1y2, W1=0y2f(x)y2.W2=y10y2f(x)
Example;
Solve y''-y'-2y=2e-x.
Solution
Form an auxiliary equation.
m2-m-2=0
m-2m+1=0
m=2, m=-1
Therefore,
yc=c1e2x+c2e-x
So that, y1=e2x, y2=e-x,f(x)=2e-x,
Next, compute the W.
We2x,e-x=e2xe-x2e2x-e-x= -3ex
W1=0e-x2e-x-e-x=-2e-2x
W2=e2x02e2x2e-x=2ex
So that,
u1'=W1W= -2e-2x-3ex=23e-3x
u2'=W2W= 2ex-3ex=-23
Therefore,
u1=-29e-2x+C1
u2=-23x+C2
So that,
yp=-29e-2x-23x
So that the full solution is
y=c1e2x+c2e-x-29e-2x-23x

b) Cauchy-Euler Equation
an(x)dnydxn+an-1(x)dn-1ydxn-1+…+a1(x)dydx+a0xy=f(x)
where the coefficients an, an-1, . . ., a0 are constants, is known diversely as a Cauchy–Eulerequation, an Euler–Cauchy equation, an Euler equation, or an equidimensional equation. The observable characteristic of this type of equation is that the degree k =n, n –1, … 1, 0 of the monomial coefficients xkmatches the order k of differentiationdkxdyk: The transformation x = etreduces the equation to a linear O.D.E. with constantcoefficient in the variable t. Notice that we assume x > 0, and t = ln x.Using chain rule, since y is function of t through x: dydt=dydx×dxdt=dydxet=dydxx

Then,
dydxx=dydt
Remark: For the second order equation, the transformation produces: ax2d2ydx2+bxdydx+cy=gx, x>0
Taking x=et
Example:
Solve x2y''-2xy'+2y=x3.x>0
Taking the transformation x=et the equation reduces to
yt''+(-2-1)yt'+2y=e3t
Or
yt''-3yt'+2y=e3t
Form an auxiliary equation.
m2-3m+2=0
m-2m-1=0
m=2, m=1
Therefore,
yc(t)=c1e2t+c2et
yp(t)=Ae3t
yp'(t)=3Ae3t
yp''(t)=9Ae3t
And substituting into the equation,
9Ae3t-3(3Ae3t)+2(Ae3t)=e3t
9Ae3t-9Ae3t+2Ae3t=e3t
2Ae3t=e3t
A=e3t2e3t=12
yp(t)=12e3t
Therefore full solution is
y=c1e2t+c2et+12e3t

PROJECT 4: BUNGEE JUMPING
PROBLEM 1
Solve the equation mx''+βx'=mg for x(t) given that, you step off the bridge that is no jumping, no diving! “Stepping off” means that the initial condition are x0=-100, x'0=0. Use mg=160, β=1, and g=32.

Solution:
mx''+βx'=mg ……………. (1)
x0=-100,
x'0=0
mg=160, β=1 , g=32
mg=160
m32=160
m=5

m=5 and β=1 into (1)
5x''+ x'=160 ……………….. (2)

The Auxialiary Equation:
5m2+m=0
m5m+1=0
m=0 , 5m+1=0
m=-15
So,
m=0 and m=-15 into x=emt
x1= c1e0t and x2= c2e-15t
x1= c1

So the general equation is
xc= c1+c2e-15t

Solve for xp
xp=At , x'p=A , x''p=0 into (2)
50+A=160
A=160
∴xp=160t
So, xt=xc+xp
xt= c1+c2e-15t+160t ………………(3)

When x0=-100 into (3)
-100= c1+c2e-15(0)+160(0)
-100= c1+c2 . . . (I)

When x'0=0 into (3)
x't= -15c2e-15t+160
0= -15c2e-15(0)+160
160= -15c2
c2=800 . . . (II)

Substituted (II) into (I)
-100=c1+800
c1=-900
∴xt= -900+800e-15t+160t

PROBLEM 2
Use the solution from problem 1 to compute the length of time you free-fall (that is, the time its take to go the...