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

ad2ydx2+(b-a)dydx+cy=get

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...