11.6 Series of Orthogonal Functions: Mean Convergence

The first and third equations are related to Legendre's equation.

This is Legendre's equation.

This type of convergence is called pointwise convergence.

This section describes a different kind of convergence that is useful for eigenfunctions.

There are a number of reasonable meanings attached to this phrase.

It is very easy to write down Eqs.

There are several deficiencies.

It isn't often useful in practical problems.

The least upper bound is smaller than any other upper bound.

If the function has one, the lub of the function is equal to its maximum.

Writing out something.

The coefficients were defined.

The coefficients (9) are the same as those in the eigenfunction series whose convergence was stated.

In two important respects, Equation (9) is noteworthy.

This is a practical significance.

The only thing required is to compute from Eq.

Mean convergence is a different type of convergence than pointwise convergence.

A series can converge in the mean.
The area between two curves, which behaves in the same way as the mean square error, may be zero even though the functions are not the same at every point.
They don't have to affect the mean square error on any finite set of points.
Even if an infinite series converges at every point, it may not converge in the mean.
The mean square error could become unbounded.

The coefficients were given by Eq.

There is a similar definition for completeness.

The convergence of series is one of the reasons for the Theorems.

There is a continuous derivative there.

Theorem 11.2.4 can be generalized if pointwise convergence is replaced by mean convergence.

The first thing we need to do is define what a square integrable function is.

The following Theorem is similar to the other one, but it involves mean convergence.

The class of functions specified in 11.6.1 is very large.

Some of the functions in the class of square integrable functions are not differentiable at any point.
The functions have mean convergent expansions in their eigenfunctions.
Many of the series do not converge pointwise.
Mean convergence is more associated with a series of functions than with pointwise convergence.

There is a special case of the general theory of Sturm-Liouville problems discussed in Chapter 10.

The series is the same as the series discussed in the section.

The error is close to 1.

Pointwise convergence is not implied by mean convergence.

There are further results associated with mean convergence.

The result is called Parseval's equation.

The numerical and graphical data about Bessel and Legendre functions can be found in the following books.

See SSM for 9.

If the initial value is negative, there are 1 detailed solutions.

The initial value is less than 5.

The solution approaches equilibrium faster in two places than in one.

-2 is not stable.

You can see SSM for 20.

3 is the number from the data.

You can see SSM for 23.

The equation is not perfect.

The equation is nothomogeneous.

Both the Legendre and Airy equations are self-adjoint.

If origin is interior to interval, it is independent.

You can see SSM for 18.

+ * * 12.

There is a singular point.

There are irregular detailed solutions.

The graph is a circle with a center at origin.

The origin is approaching the graph.

They are linearly independent on every interval.

It's the natural frequency.

SSM has a solution to five digits.

For detailed solutions, see SSM.

As the initial point moves away from the equilibrium point, the period increases slowly.

You can see SSM for 25.

You can see SSM for 1

No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or108 of the 1976 United States Copyright Act.

The Permissions Department of John Wiley & Sons, Inc. is located at 605 Third Avenue in New York.

The seventh edition of Eleanor Deviations and Boundary Value Problems and Eleanor Differential Deviations are included in this supplement.

There are problems from each section of the text in the supplement.
The complete details in determining the solutions are usually the only information given.
In order to give the student a complete set of examples from which to learn, the problems chosen in each section represent the variety of applications and types of examples that are covered in the written material of the text.

Don't use the guide to solve the problem.

Look at the beginning of the guide and try again to get the complete solution.

Since there are more than one correct way to solve a problem, you should compare your final solution with the one provided by the guide.

Why was that problem assigned?

The details in the solutions manual are important for the student's understanding of the underlying mathematical principles and applications, even though the use of a symbolic computational software package would greatly simplify finding the solution to a given problem.

In some cases, the software packages are necessary for completing the given problem, as the calculations would be overwhelming using analytical techniques.
In these cases, some steps or hints are given and then reference is made to the use of an appropriate software package.

The following abbreviations have been used to simplify the text.

I would like to thank Mrs. Susan A. Hickey for her excellent work on the manuscript.

Dr. Torok has provided assistance with the preparation of the figures as well as assistance with many of the solutions involving the use of symbolic computational software.

There is an increase in y(t) when we see that yC/ > 0 is increasing.

The decrease in y(t) is due to the fact that we have yC/ 0 and thus y(t) is decreasing there.

Y(t) is different from 3/2 as tAE*.

The direction field shows that the solution is decreasing here.

We have yC/>0 and thus y(t) is increasing here.

As the slopes get closer to zero, we conclude that yAE-1/2 is tAE*.

We have equilibrium solutions for the numbers 0 and 4.

If the solution approaches 4 from above, it's for y > 4, yC/ 0 If 0 y(0) 4, then yC/ > 0 and the solutions grow to y as tAE*.
The solutions differ from 0 for y(0) 0.

The equilibrium solution is yC/ > 0 for all y.

If y(0) > 0, solutions will differ from 0 and if y(0) 0, solutions will differ from tAE*.

All solutions approach the equilibrium solution if qC/ is 0 or c is 10gm.

The solution starts above the equilibrium solution and decreases as it approaches 5/2 as tAE*.

The solution starts below 5/2 and grows rapidly and approaches 5/2 from below as tAE*.

The solution to Eq(i) is y1(t) + b/a.

If p0 is the same as p0, then c is the same as p0 and thus p is 900.

If p0 900, this decreases, so if we solve for T, we get eT/2, or T is 2ln[900/(900-p0)].

The yield is -r or ln Q.

Q Q is ce-rt and Q is Q.

The 100e-r is set at 82.06, which yields a r of.1980/wk or r of.02828/day.

Rewriting the D.E.

Q(t) is CV and thus Q(t) is Th D.

CV is 20, CR is 2.5 13c.

This is the third order.

The linearity of the D.E.

is unaffected by the terms t and cos t.

We have 9e for y (t) and -3e for y.

If f(t) is the same as U f(s)ds, then UC/(t) is the same.

Substitute into the D.E.

If y is to be a solution of the D.E.

substituting into the D.E.

It is true for all t and x.

te3t + et is the same as e3t(yC/+3y) or e3t(t+e-2t)

The evaluation of Ute3tdt is based on integration by parts.

The equation needs to be divided by t in order to form Eq.

If you want to integrate the right side, you can use an integral table, or use a symbolic computational software program.

To find the value for c, set t to 0 in y and 1 in the initial value of y.

The solution of the initial value problem is given by -2+c.

Settingt is 1 and y is 1/2.

tsint and t2y are set at p/2 and y is set at p/2.

We must have c if y(t) is to remain bound.

Since both l and are positive numbers, lim y is zero.

There isn't a unique answer for this situation.

Thus yC/ + 2y is 5 - 2t.

The method of variation of parameters for the simplest case is shown in this problem.

Assume A(t)exp(-U)dt is A(t)e2t.

Differentiating y(t) and substituting into the D.E.

A(t) is a combination of t3/3 and c, which yields the solution.

Problems 1 through 20 are the same as the examples worked in this section.

The first eight problems do not have I.C.

3y + y + x + c is the result.

If you integrate each side you get arcsiny of lnx+c, so y is sin[ln|x|+c], x p 0 Since both sides are zero, y is +- 1 and satisfy the D.E.

The negative square root has to be used.

The solution is valid only for the open interval, since yC/ does not exist for x or x.

The solution is valid for all x.

Adding 1/3 to both sides will complete the square on the left side.

There is a chance that x - 15 /2 is discarded.

Separating variables gives y - 5y, then y - 1y, then y - 2y, then y - 3y, then y - 4y, then y - 5y, then c.

The interval of definition is found numerically.

The approximate interval is 2.95 PS x PS 2.1863, or x-p/2 PS 0.6155, if you take the indicated square roots.

3y - 6y is 0, or y is 2.

The solution is defined by finding the zeros of the denominator and has vertical asymptotes at the end points of the interval.

The initial point is -.126 x 1.697, found by solving sin2x is -.25 for x and noting x is the initial point.

The solution clearly shows that the critical point is a maximum.

We note that yC/ 0 for y > 4 and yC/ approaches zero as y approaches 4 by sketching the direction field and using the D.E.

For 0 y 4, yC/ > 0 and again approaches zero.
lim y is 4 if y is 0.

C and ce are related.

Setting this equal to 3.98 and solving for t yields t.

Separating variables yields a result.

A a(ay+b) yields the answer.

The last equation in (c) has a different ending.

The D.E.

is simplified on the right side.

There are two solutions to this equation.

For v p 1, the variables are separated.

In Prob.33, substituting y into the D.E.

The answer can be obtained using partial fractions.

It can be verified that both forms satisfy the D.E.
by differentiating them.

q(t) represents the amount of dye present at any time.

The original amount of salt in the tank should be 0, the amount of salt entering per minute should be 2, and the amount of salt leaving per minute should be 2.

Thus dS/dt is 2 - 2S/ 120.

After 10 minutes, we need to find the amount of salt.

For the first 10 minutes, if we let Q(t) be the amount of salt in the tank, it would be 2, 2, 2.
The new I.V.P.
was created because no more salt is allowed to enter.
The solution of the problem is P(10), which yields 7.42 lbs.

Salt flows into the tank at a rapid rate.

Set is zero in Eq.

Set r and t and 40 and S(t) and $1,000,000 and solve for k.

Set k, t, 40 and S(t) are all equal to $1,000,000 and then solve numerically for r.

The rate of accumulate due to interest is.1S and the rate of decrease is k dollars per year.

S(t) is 8000 - 10k(-1).
Setting S is equivalent to substitution of t is equivalent to $3,086.64 per year.
Over the course of 3 years, $9,259.92 has been paid in interest.

If we assume continuity, we can either convert the monthly payment into an annual payment or the yearly interest rate into a monthly interest rate.

Proceed as if it were Prob.

The Eq is being used.

An integrating factor and integration by parts can be used.
Setting S(t) 0 and solving numerically for t yields 135.363 months.

The problem can be done numerically.

We have (.1.2sint)dt by separating variables and y(t) cexp.

If y(0) is the same as y(t) exp.

The doubling time has not changed because of the same numerical equation.

If (T) is 15 we get 7 and 13 T yields ln(8/13)/-ln(13/12), which is ln(13/8)/ln(13/12) min.

Then the concentration is given by x(t) + Q(t)/ 1200.

In the solution found in (a), set k, P, t, and c(T) to zero.

If we measure x upward from the ground, then Eq.

30 and x(t) are 20t - (g/2)t + 30.

At the ground, x(t) is 0 and thus 20t - 4.9t + 30 is 0.

To find the distance when the parachute opens, set t to 10.

The parachute opens the I.V.P.

The limiting velocity comes from letting tAE*.

After the parachute opens, the length of time the skydiver is in the air will be calculated.

If x is measured upward, then Eq.

Sect.

1.1 becomes m.

26b.From part (a) v(t) + [v + ]e

The answer is given when you solve for e.

This problem is the same as example 4.

The escape velocity for a surface launch is 2gR.

The time is right for the ball to reach the wall.

If you plot the left side as a function of A, you can find where the zero crossings are.

A equality sign since we want to clear the wall.

We get U as a result of Solving for U.

When the denominator has a maximum, 175sin2A-7cos A have a minimum.

350cos2A + 7sin2A is equivalent to 0 or tan2A is equivalent to -50.

The equation can be written in Eq.

The solution will be continuous on 0 t 3 since the initial point is t.

The numbers are -2t +c or y.

From the direction field to the D.E.

1/2 22a.

Setting t is 2 in y and y is -1, as required.

Take the derivative and substitute into the D.E.

to show that f(t) is a solution of the D.E.

1(t) + p(t)y1(t) + y2(t) + p(t)y2(t)

A linear D.E.

is given by setting v and y together.

This will be put into the D.E.

The linear D.E.

is given by v.

The solution is yielded by setting v to 1/y.

We can solve the I.V.P.

if g(t) is continuous on the interval 0 PS t PS 1.

Problems 1 through 13 follow the pattern illustrated in Figure2.5.3 and the discussion following Eq.

The critical points are y and x.

The graph is positive for O y 1 and 2 and negative for 1 y 2.

There are variables to get.

At that point, the solution has a discontinuity.

The result will be fC/(y ) > 0.

The desired solution can be found by solving for r.

The equilibrium is given by k.

The equilibrium height needs to be less than h.

Use the quadratic formula to solve for y.

There are no critical points if h > rK/4.

The equations give the I.V.P.

There are separate variables to get.

The answer is given by integration and solving.

If you plot dx/dt vs x, you'll see that p and x are critical points.

It's important to note that min(p,q) and max(p,q) are both 0 for x between min(p,q) and max(p,q).
x is stable while x is unstable.

Solving for x yields a solution.

x(t) approaches p as tAE * To integrate and solve the D.E.

We get y(x,y) with respect to x, which is x - x y + 2x + H.

hC/(y) is 6y + 3 and h(y) is 2y + 3y.

If you substitute h(y) into y(x,y), remember that the equation which defines y(x) is y(x,y)

The equation that yields the solution is x - x y + 2x + 2y + 3y.

The equation is written in the form M(x,y)dx + N(x,y)dy.

The equation is exact.

M(x,y) with respect to x yields y(x,y)

The D.E.

is the abbreviation for M (x,y) and 2sinx.

If you want to find y(x,y), you have to integrate by parts.

The equation xdx + ydy is given.

The D.E.

is the number of M and N.

The negative square root is used in order to satisfy the I.C.

3x + 2xy is the number.

The number of D.E.

It's easy to get 2 x N(x,y) + 2 x (x,y) + 2 x siny + g(x)) from this equation.

The implicit solution is given for g(x).

The problem is similar to the one leading up to Eq.

We find from Eq that m depends on y.

This is the last D.E.

An alternative approach to Problems 27 through 30 is provided by this method.

The integrating factor is 3m.

Thus adding the D.E.

The last equation can be changed to ysiny dy.

The first and last terms can be integrated.

We have xy + ycosy.

The formula is yn + h (2yn - tn + 1/2) for n, t0 and y0 Thus y1 +.1(2y0 - t0 + 1/2) is 1.2, y2 + 1.25 is 1.2, and y3 + 1.54 is 1.54.

Only results for n, 3 and 7 are needed to compare with part a.

Again, use the same formula with h and n. Only results for n are needed to compare with parts a and b.

If you want to get (e-2ty)C/, you need to rewrite the equation in the form yC/ - 2y.

We get ce-2t + t/2 with the integration of the right side by parts.

The solutions seem to converge for y(0).

The solutions seem to differ.

For h, we get y1 being -2 +.1(4/5) and y2 being -1.92 +.1( )

1 + (1.617043) 2 Thus, y(.5)

2 y2 is -.75 +.1

The large differences are explained by two factors.

The slope of y, yC/, becomes large for values of y near -1.155 from the limit.

The slope changes sign is -1.155.
The slope yC/ here is large and positive, creating a large change in y8 as part a.
A large negative slope is created by part b, y(1.65) at -1.125.
The slope at this point is positive and the rest of the solutions grow to -3.098 for the approcimation to y.

For the four step sizes, the approximte values are 3.5078, 4.2013, 4.8004 and 5.3428.

Since these changes are still large, it is hard to give an estimate other than y(.8) is at least 5.3428.
Further approximate values of y(.8) are found by using h with.005,............

The better estimate is between 5.8 and 6.

There is no reliable estimate for y(1), which is consistent with the direction field of Prob.9.

This term is positive and grows very large for tn > 2.

The term decreases and becomes negative for tn at 1.6 for h.

y(2.00) has the approximations of.1,.05 and.01.

The Eq is being used.

Setting n + 1 becomes yk for 1 + 2h for k. Since y0 is 1 we have y1 and y2 being 1 and 2h.

Continuing in this fashion, we get k/2 + 1/2.
Pick h if you want x/k.
Substitute for h in the last formula to get yk.

When t is 1 and y is 2, we have s and w being equal.

The first example of the text is from Eq.

1 +f(t) is the number.

F(t) is equal to e-1.

To verify this form for f is our hypothesis.

The Eq is being used.

Where I mean k-1.

From the equation we have y.

n(y -12) + 12

The 1-rn solution is rny.

You must solve the problem.

Substituting v into the equation.

An initial value of.2 was chosen for both parts of the problem.

A slightly different number of iterations is needed to reach the limiting value for different initial values and computer programs.

The limit of.65517 is reached after approximately 100 iterations.

The limit of.66101 is reached after approximately 200 iterations.

The limit of.6555 is reached after around

The solution is between.63285 and.69938 after 400 attempts.

The solution is between.73770 and.59016 after approximately 130 attempts.
The solution can be found between.55801 and.76457.
The oscillations were verified for each case.

For an initial value of.2 and r of 3.448, we have the solution between.403086 and................

The eighth decimal place is not fixed after 3570.

The solution is dependent on the four values:.
3.45,......................

Before trying to find a solution.

The student needs to classify the D.E.

2yu is a linear expression of U(y).

It depends on x and 12.

The integrating factor is dependent on 20.

It depends on x and x only.

Integrating factor is dependent on y.

Assume y is the same as yC and y is the same as y.

6r -r-1 is 0, or 3r-1 is 2.

The roots of the equation are r + 5r, and the characteristic equation is r + 5r.

The characteristic equation is r - 9r + 9 - 0 so that it is used in the formula.

c exp[(9+3 5 )t/2] + c exp[(9-3 5 )t/2

Substituting y is done in the D.E.

Since e has a positive correlation for t > 1, y.

The characteristic equation is (r-2), or r + r - 6.

The solution is for the D.E.

ln2 is 4 ort.

The minimum point is indicated by the second derivative test or a graph of the solution.

The general solution is c e + c e. The I.C.

is being used.

The second term becomes unbounded, but not the first, so there are no values for which all solutions become unbounded.

The solution decays to zero if b is 2 and y is e. There is no minimum point for the solution for b > 2.

The D.E.

is "let v be y', then v' be y"

Set v to yC/, then vC to y, and finally the D.E.

If c is greater than 0, let it be k.

If c is 0, let it be -k.

Then yC/ is 2/(k + t) so that y is (2/k)tan (t/k) + c.

y is a solution of the D.E.

Following the procedure outlined, let's say v is dy/dt and y is dv/dt.

2y v dv/dy + 2yv is the number.

Section 3.2 is written in the form +- ydy/(y+c) 1/2.

If v is yC/, then v is vdv/dy and the D.E.

We get a separable equation in y and t, which is solved to yield the solution.

Both t and y are missing so either approach will work.

The variable is missing.

The D.E.
is the number of vC/, then vC/ and the number of y.

The last equation should be solved and applied to the I.C.

Cos t and sin t are equal.

W(x,xe) is the same as xe and xe is the same as xe.

The only point of discontinuity is t. Since the initial point is t, the largest interval is t.

Since the D.E., Theorem 3.2.2 is not contradicted.

L is a linear operator.

g(t) p 0 is the same as g(t) if and only if c is 1.
This is not a contradiction of the Theorem.

W(f,g) is either tgC/ or gC/.

Suppose that P(x)y is + Q(x)yC is + R(x)y.

On expanding the right side, we find that f'(x) is R(x) and P'(x) is Q(x).
The necessary condition P"(x) - Q'(x) + R(x) is satisfied if R(x) is Q'(x) - P"(x)

The condition for exactness is satisfied if P(x) is x, Q(x) is cosx, and R(x) is sinx.

From Problem 27 f(x) is Q(x) - P'(x) is cosx -1, so the D.E.

m(x)P(x)y + m(x)Q(x)y' + m(x)R(x)y is what we want.

To equate coefficients of y, y' and y, expand the right side.

This gives m'(x)P(x) + m(x)P'(x) + f(x) Eliminate f'(x) to get the adjoint equation Pm" + (2P' - Q)m' + (P" - Q' + R)m.

R is a(a+1) and P is 1-x.

2PC/ - Q is -4x + 2x is -2x is Q and P is R.

Y Py" + (2P' - Q)y' + (P" - Q' + R)y

substituting for P, Q, and R, we get Py" + Qy' + R. The equation is the same as before.

Pm is the adjoint of Py and Qy' and R is the adjoint of Q and P. The two equations are the same if P' is Q.

P'(x) is 2x and Q'(x) is x.

The Bessel equation of order n is not self-adjoint.

W(t,t ) is -2/t p 0.

For t and hence f(t) - 3g(t) - 0 for all.

The D.E.

is written.

W(y,y) is the same as c/t.

W(y,y )(1) is 2 and thus W(y,y )(5) is 2/25.

The point in I where both y and y disappear.

Then W(y,y )(c) is the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the

Let -1, t, and p.

Clearly, they are solutions.

On 0 t, f(t) and g(t) are equal.

0 t 1 and 0 t 0.

This will be put into the D.E.

Rewriting the D.E.

y + c sinx + x + lnt, t > 0.

If r is the same as c then y(t) is the same as c.

Let the number be v/t.

Substituting in the D.E.

Thus v is a combination of c and lnt.

Substituting y is done in the D.E.

This is the first order.

There are two possibilities in this case.

Substituting Z into the D.E.

We assumed Y was A + Bsin2t + Ccos2t.

Substituting in the D.E.

The D.E.

needs to be solved first.

Y is for Acosht + Bsinht.

This will be done in the D.E.

Substituting Y in the D.E.

Y is 4Ae +4(At+B)e. Substituting into the D.E.

First you have to solve the I.V.P.

Substituting in the D.E.

Y - Y is a solution of ay, byC, and cy.

We wrote the D.E.

from Problem 32.

D.E., we assume Y is the cost.

The solutions of the D.E.

are thatt and te.

Y + te U is a combination of tu and te.

Limits are put on the integrals.

Clearly Y(t) is 0 and YC/(t) is 0.

We put the D.E.

first.

2 and d are tan-1(- 3 ) and p is 2p/3.

The motion is vibrating.

14t is either p or t.

We use it.

The spring constant is 3920 dyne/ cm.

The I.C.

is 6 t.

T is 2p 14 and p is 7.

4+25/6 e-10t .05 yields t.

6 t-d is.4086 for t.

The values for L, C and R were given.

.2Q is a combination of 3x101 QC and 105 Q.

QC/(0) is the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of

64 lbs/ft.

is the 8/1-8 ratio.

U can disappear at will.

The static case should be considered first.

The positive direction leads us downward.

Bsin2t is 4cos2t.

Referring to something.

We have to solve the three I.V.P.

1.5468 x 10 coulombs is the 4e + 3 equation.

This D.E.

is compared to this D.E.

Rsinq is 2sinq.

The equation is r -5r + 3r + 1

Using the procedure suggested.

The general solution is given by the teacher.

The D.E.

needs to be solved first.

There is no need to change this choice.

A is 1/2 and B is 3.

The desired solution is given by D.

Applying the I.C.

Y (t) is the same as A t + A t.

Y (t) is the same as A (t) and A (t) and A (t) and A (t) and A (t) and A (t) and A (t) and A (t) and A (t) and A (t) and

A t + A t + B )cost + (C t + C )sint.

D(D-1) 2y is 2e.

The right side of the D.E.

is destroyed.

The cost is 0 (b).

Y(t) is a combination of cost and sint.

W(t) + cexp(-Up (t)dt) equals ce.

W(t) is 2e.

The series converges for x 2.

For the problem f(x) is sinx.

F(x) is the number for this problem.

For the problem of f(x) + lnx.

Na ]x.

-2 a 3!...

For n to be 1,2,3,...

Substituting in the D.E.

Substituting in the D.E.

Substituting in the D.E.

A second solution can be obtained by selecting a value of 0.

There is a [(x-1) - (x-1) 4/4 - (x-1) 5-20 + (x-1) 7/28 +...])

y'(0) is a 0 and y'(0) is a 1

Thus phiv(0) is - 0 - 3(-1).

The zeros of P(x) are 2x and 3x.

x has a zero at x and r is the number.

20a x 30a x 42a x

A /504,...

A for n is 3,4

x is 0.

The equation can be written.

The D.E.

is written.

The D.E.

is written.

x +-1 are singular points.

The D.E.

is written.

The D.E.

is written.

Substituting in the D.E.

The point at zero is a singular point.

The D.E.

was compared to the D.E.

If y is x then r(r-1) + 3r + 5 are equal.

The solution of the D.E.

is 0.

We have F(r) as r(r-1), 3r, 5, and 2r.

The roots of F(r) is given by Eq.

Y is assumed to be v(x)x 1.

V''(x)x 1.

Substituting in the D.E.

The D.E.

is transformed by the change of variable x.

The D.E.

is being put in.

n is 0,1,2 and a is a.

,n,n,n,n,n,n,n,n,n,n,n,n,n,n,n,n,n,n,n

Since a negative number it follows that a negative number is a negative number.

The D.E.

is being put in.

The number is 0.

Also note that J is 0 as x is 0.

Q(x) is 2x and R(x) is 6e.

This series is being put in the D.E.

We have set a value of 1.

These are replaced in the D.E.

We start writing the D.E.

1 (x-1) 2 +...

We should write the D.E.

We need ara of 0 which requires r of 0.

It's clear that x is a singular point.

We use Eqs.

It's clear that x is a singular point.

We have to calculate a' (-1).

2 and G' (-1)/G (-1) are related.

To compare with the D.E.

We have to verify that J (l x) complies with the D.E.

J' (t) + J' (t) equals 0.

Let's say l x.

In the given D.E., substituting y is done.

J' (t) + J' (t)

The given D.E.

is a solution of y and j.

There is a graph of f(t).

PStsinat is 2as/(s +a ) 2, s > 0.

Take the Laplace transform of the D.E.

The I.C.

is being used.

The pattern of Eq.

is followed.

4Y(s) + 4Y(s) equals 0.

The I.C.
is being used.

+Y(s) equals 0.

The I.C.
is being used.

2 + 1 + 7 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +

Since lim (sint)/t is 1, f is continuous.

PSt e is 2 and 3.

Taking the Laplace transform of the D.E.

Y(s) - sy(0) - yC/(0) + YC/(s)

Y is able to satisfy YC/ + s Y.

The result follows.

PSu (t)g(t-1), e PSg(t), is 2/s + 1/s.

H(s)

A(-1 k e /s) + A(-e ) k

U e f(t)dt is defined as PSf(t).

periodic with period 2 is the function f.

Problem 28 gives us PSf(t)

1 + sint - cost goes up.

The I.C.

was introduced.

Applying the I.C.

to the Laplace transform.

If g(t) is sint, then up(t)sint is sint + up(t)sin.

Taking the Laplace transform of the D.E.

The plot will be zero in all cases.

The graph on 0 PS t 6p is dependent on how large n is.

Taking the Laplace transform of the D.E.

The cost and kup are different for nAE*.

The Frequency is 2p from the graph of part a.

The I.C.

is used to take the Laplace transform.

Taking the Laplace transform of the D.E.

There is a solution for Problem 7.

Up(t)sin(t-p) + U2p(t)sin(t-2p) + o + U10psin(t-10)

Taking the transform and using the I.C.

The format of Eqs is being used.

We can define g(t) as t and h(t) as cos2t.

G(s) is the number of PSg(t) and H(s) is the number of PSh(t)

The D.E.

has a Laplace transform.

The I.C.

is used to take the Laplace transform.

Let x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x

The first D.E.

was solved.

The I.C.

is being used.

There is a graph on the right.

The first D.E.

was solved.

Then use the new ones.

The Eq is being used.

The second row should be added to the first row.

The second row should be added to the first row.

The second row should be added to the third row.

The second row should be added to the third.

The third row is equivalent to 1/3.

2 1 1 is the augmented matrix.

From the third row we have a zero.

The equation c + 4c is equivalent to the third row.

0 t t t

The eigenvector is related to l.

The numbers are -l3 + 6l2 + 15l.

x2 is equal to 0.

C + e-2t.

c1 -4 et + c2 -1 e-2t + c3 2 e3t.

The general solution of the D.E.

c2 and c1 are equal.

r2 + 2r + 1 is a value.

The desired bifurcation point is here.

2 is able to satisfy 2i and 2i is able to satisfy 2i and 2i is able to satisfy 2i and 2i is able to satisfy 2i and 2i is able to satisfy 2i and 2i is able to satisfy 2i and 2i is able to

x1 is 0 and ix2 is 0.

r2 + r/2 + 17/16 is 0.

The critical value of a yield is r1 or r2

There are two important values.

2 ix1 + x2 is equal to 0.

2 isin and 2cos are used for t > 0.

The values are given by (r+1/4)[(r+1/4) 2 + 1])

satisfying the I.C.

It is called Ps-1(0).

Ps that satisfy the I.C.

c1 is 0 and c2 is 1/2.

The third column is called T.

The D.E.

has one solution.

Which confirms Eq.

The equations yield e1 - 2e2

There is a graph on the right.

1 is attracted by the origin.

R is the eigenvalues.

The given D.E.

has one solution.

This expression will be put into the D.E.

One solution of the D.E.

2xte2t + xe2t + 2ee2t.

Ps -1(t)g(t) from Eq.

There are 27 words.

The desired solution is yielded by Ps(t).

The formula is yn+ h+ tn+ yn+.

We find yn+1, which is yn + h (3 + tn))/ (1+n).

There is a local truncation error.

PS, hL,En, and max are a part of the equation.

The rest follows from Eq.

We have en +f +(tn)h2/2.

2 and k3 are the same as yn and tn.

F(tn,yn) is equal to ft(tn,yn) + fy(tn,yn)f.

Substituting this in the book.

1.57574 )

We use it.

Let P2(t) be At2 + Bt + C.

Subtracting from the total.

A is 1/6, B is 1/6.

The Eq is being used.

2 y1 (.1/6).

The family of lines is not related to c2.

The lines are vertical.

r2 - (a11+a22)r + a11a22 - a21a12 is 0.

The middle quotient is ignored.

The ellipse is provided by B2 - 4AC 0.

The approach indicated in Eq.

is being utilized.

r2 + 1 is either 0 or +- i.

The desired solution is -2)

If that is the case, imagine that t1 > t0 Let's say t1 - t0

G[f(t+T), y(t+T), G[F(t), Y(t)]) is the yC/(t)

The results are summarized in a table.

R is 1

This can be written in the form of Eq.

The table has an unstable saddle point.

For x p 0, we have a dy/dx of (dy/dt)/ (-2y+x3)/x.

We estimate vc using several values.

The motion is damped about x.

Setting c was 0 in the equation.

L/2g dq/ cosq gives dt.

One eigenvector is.

The second eigenvector is.

If (x,y) starts in the first quadrant, it's 6f.tAE.

r2 + 4r + 4a + 0, or r2 + 2 a.

It is also an unstable saddle point.

The ratio is cK/g/ ac K/a.

Vy is 2cy and thus Eq.

2cy(-2x2y - y3) is a combination of 2ax4 and 2c-a.

V is a negative definite.

.V(x,y) is the number of 2ax(-x3 + 2y3) and 2cy(-2xy2).

The correct system is dy/dt.

g(x)y + y[-g(x)] is 0.

V is positive definite.

2xy - 2y2 - 2ysinx is 2y[-y + (x - sinx)])

Then V(x,y) > 0.

V is for g(x)y + y[-c(x)y - g(x)]

D 0 and A 0.

It follows that 4AC - B2 > 0.

.V[x(r,q), y(r,q)) PS -r2 + 2Mr(er) + Mr.

The direction of motion is in the opposite direction.

Fx(x,y) + Gy(x,y) is 3+4x2+y2.

Since x and y are related, y is a solution of Eqs.

fC/(t)yC/(t)dt -yC/(t)fC/(t)dt is 0.

Setting x' is equivalent to 0 and solving for y is equivalent to x3/3.

A is the same as A.

A is the same as A.

If r is 28 then Eq.

-2a[r21 + 2ar1 + (a2+b2)]) is equal to 0.

A comparison of Eq.

2s-[x2(r+1)xy+y2]-bz2

Writing V is done in the form of Eq.

If we choose c, that's it.

There are several cases shown.

The values of T are looked for by us.

You can see the next page.

The given f(x) has a Fourier series.

Substituting for f(x).

These values will be put into Eq.

L is the desired solution.

Substituting for f(x).

Using equations.

For this part, we will assume f(x) is continuous.

n is bound as n. ].

Both n2an and n2bn are bound.

e-x is neither odd nor even.

Applying Eqs.

The Eqs are used for the sine series.

Using equations.

We use Eqs since we want a sine series.

Property 3 is the same as Property 3.

F(x) is an odd function.

Since f(x) is equal to x and bn is equal to 2L.

We are looking for solutions of the form X(x)T(t).

Substituting into the P.D.E.

L(X' + X) is 0 and T' + lT is 0.

We are looking for solutions of the form X(x)Y(y).

Substituting into the P.D.E.

The procedures of Eqs are followed.

The constant is X''X, which is 4T'/T.

X(x) is the number of C1coslx + C2sinlx and X(0).

X(x) is not acceptable.

X(x) is the number of the sin, n is the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of the number of

400ln(80/p)/p2(.86) is the total time.

Substituting x,y,t in the P.D.E.

T equals 0.

X''/X is s1 and Y'' is s2 so X'' is 0 and Y'' is 0.

T' + a2l2T is 0.

Again, v(x) must satisfy v'', v'', v'', v', v', v', v', v', v', v', v', v', v', v', v', v'

V(x) is equal to T(x+1)/L+1.

Using equations.

The yield is 160.30 seconds.

Substituting the I.C.

25cos dx is related to sin.

Substituting X(x)T(t) into Eq.

X'' - sX and T' - a2sT.

X'(L) is equal to 0.

X(x) is the number of k1sinlx + k2coslx.

X(0) is 0 and k2 is 0 and thus X(x) is k1sinlx.

Differentiating setting x and L yields the same result.

It is necessary to satisfy the I.C.

To get the solution you want.

Substituting f(x) into Eq.

The graphs in part b can be understood.

The motion is governed by the law.

Substituting this in the book.

T'' + a2sT equals 0.

Applying the B.C.

-2aph(x) is from the second equation.

Substituting R(r)Th(th)T(t) into the P.D.E.

R'/R + R'/rR + Th''/Thr2 equals T''/a2T.

The two O.D.E.

are assumed to be led by X(x)Y(y).

X'' - sX is 0, Y'' is sY is 0.

solve the D.E.

C1sinhly + c2coshly are the numbers.

Using the B.C.

The Eq is being used.

0 is 2p.

Th(th) is a combination of c1coslth and c2sinlth.

These values will be put into Eq.

R(r) is the number of k1rnp/a and k2r-np/a.

X(x) is equal to sin and l2 is equal to 2.

C1exp[npy/a] + c2exp

X'' - sX is 0, Y'' is sY is 0.

The value of Y'(b) is 0.

Y'' + sY is zero with these B.C.

X(x) is the number of Asinh and Bcosh.

To satisfy the remaining B.C.

The coefficients are determined by B.C.

If l is 0, set -l to get y.

The ph and sinu are related.

2s' + 4s are 0 and the D.E.

is e-2x.

L is 1/2.

The eigenfuction is ph 0(x) xe-2x.

2c + c 2l - (c + LCS 2l-1) is equal to 0.

If l is 0, then y(x) is c1 + c2x and the B.C.

The D.E.

is being used.

There is a note that says ph (x) is 1 satisifes Eq.

The Eq is being used.

A similar result can be obtained if b is 0.

U(x) + iV(x) is used in Eq.

L[U(x) + iV(x)]) is the abbreviation for L.

U and V are functions.

There are no eigenfuctions for l.

The D.E.

is written.

v'(1) is a + 1 and v'(1) is a a.

2a + 1 is equivalent to -2 or v(x) is equivalent to 1 - 3x/2.

F(x,t) is the same as -x.

These values will be put into Eq.

2 [-(-1)n]/np.

Y(x) is a combination of x and xu.

We have y(0) and c1 in part.

Y [py' ]' - y [py' ]' - y [q(x)y] - y [q(x)y]

p(x)W(y,y )(x)

In general, c cosx + c sinx is what y(x) is.

Setting c is equivalent to sin.

We allowed L[y] to be -(xy')' for other n and m.

L[y] is equal to -[(1-x2y']'.

Y(0) is the number we have since U(x,0) was zero.

Uxx + uyy is equal to 0.

U(0,e) and U2(e) are both zeros.

This problem is the same as Problem 22.

Th'' + n2Th + Z'' - l2Z is zero.

Z is a combination of k e-lz and k elz.

The periodic of period 2p implies l2 and n2.

d cosnth + d sinnth, n is 0,1,2...

To satisfy B.C.

The lim Sn(x) is 0 for all x.

Using equations.

The result is followed.

This result is from part a and part c.

All rights belong to the person.

The software should be easy to use.

There are clearly marked Nav igational paths.

You can use the Architect tool to run your own experiments.

Most will work through the multimedia modules.

There are up to four submodules in each module, and they range from straight forward to the advanced.

The pertinent equations and parameters will be entered into the equation of the quadrant of the tool.

The expert mode gives access to more advanced features.

The state variables can be defined as auxiliary functions.

The scales can be set manually or automatically.

The solutions' numerical values can be seen in tabular form.

A wide range of topics from mathematics, physics, chemistry, population biology, and epidemiology are covered in ODE files.

Both teams worked together to create an attractive multimedia software package.

Many people who read chapters, tested modules, and commented favorably on what they experienced, including Treasa Sweek.

The work on ODE Architect and its Companion book was supported by the National Science Foundation.

Once students understand how to deal with two-dimensional systems graphically, any of these modules/chapters are easily accessible.

There are different ways to access the modules.

This module is different from the others in that it is not divided into submod ules and that it provides a guide for learning how to navigate ODE Architect.

The Seismograph is a real world application.

The roles of eigenvalues and eigenvectors are explained.

The goal is to use graphical solutions to make handling linear systems easy.

The module and chapter give a discussion of the biology behind the models.

The pendu lum is explored using the pendulum submodule.

The three submodules of this unit tell a story and show a theorem.

Stable objects are described.

This module can be used in a liberal arts course.

The levels of the submodules can be found in the chart below.

There is a guide on the CD-ROM.

The canneries were also affected.

The California legislature imposed a moratorium on fishing for the Pacific sardine in 1967.

The decline of the Pacific sar dine population was caused by over-harvesting from 1941 to 1951.

There are long-range cycles of Pacific sardine abun dance.

To be consistent with the data and estimates used, we chose to define sardine biomass in million tons, rather than the number of fish.

If additional parameters are needed, they can be added.

The solution is displayed in a graphical representation.

We can look at the impact of harvesting.

There are four quadrants on the screen.

There should be no right-hand quadrants.

Scales appear in the two plot quadrants.

Initial conditions are set.

To set the initial conditions, go to the lower left corner.

The time interval is set.

Figure 1.3 shows the growth of the sardine population.

There is a curve in the upper right corner.

You might think that the answer to these questions is simple.

The unconstrained solution of ODE (1) is not useful over its whole domain.

The proportional growth property is not shown by ODE (2).

Exceeded carrying capacity.

The populations approach carrying capacity.

To scale axes.

The plot scales a window.

ODE Architect makes a lot of runs.

The figure shows how to set up the sweep for sardine.

The sardine has a rescaled vertical axis.

We'll look at the properties of the model we created.

The value is very close to 1.

Let's see if this improves the model.

To start.

There is a table called HTABLE.

The harvest is defined.

Let's look at the data.

The figure is a representation of the model sardine biomass.

There are laws that govern the variables.

The laws should be translated into equations.

The equations need to be solved.

The solutions should be tested in the context of the natural environment.

The model should be refined until it predicts the data.

To observe the relative stabilization of the population, re-solve 0 to 6.

The constant harvest was set at 2.71 million tons.

A constant effort harvesting function is used.

Go for a run.

Take your results and summarize them.

To demonstrate that the function behaves as claimed, use some algebra.

The function is being tested.

The growth model was used earlier.

The harvest experiments should be repeated.

The optimal harvest rate depends on the population IC.

Ordinary differential equations can be used to predict the behavior of natural processes.

Slope fields are used to guide solution curves.

ODEs modeled the path of a juggler's ball and the descent of a sky diver.

A function is a solution of an ODE if it is true.

ODE (1) has many solutions.

A particular solution is 2 + 3.

There are two techniques described in the section "Finding a Solution Formula".

An ODE can have many solutions.

An ODE has a lot of solutions.

There are laws that govern the variables.

The laws should be translated into equations.

The equations need to be solved.

The solutions should be tested in the context of the natural environment.

The model should be refined until it predicts the data.

All of the models we consider involve ODEs.

You can observe the modeling process.

You might think that the path of a sky diver in free fall looks like the downward path of the ball in the simplest juggler problem of vertical motion.

There are six sky divers in free fall from 13,500 ft.

Exploration 2.4, Problem 3 is a good place to start.

Let's look at the ODE Architect.

In this Exploration, you will see some exceptions, but usually an IVP has a single solution.

Screen 2.2 has the following ODEs.

The initial speed should be increased.

To find the ranges.

The juggler might raise his hand one foot higher.

There are two balls.

A falling body has a terminal speed.

A model for a parachute that opens in 3 seconds is needed.

The chute opens.

The faster the object cools, the hotter it is.

There is a law of cooling.

The general solution of Equation (1) has an arbitrary constant.

It was easy to find the general solution formula.

To carry out the integration, use a table of integrals.

Let's build a model that shows a room cooled by an air conditioner.

It is very difficult to solve a problem by hand.

The room is cooling.

The room temperature goes up to 80*F again.

When the step functions turn on and off, the architect won't notice.

There is a law of cooling.

T.K.

wrote "Estimating the Time of Death".

Measure the temperature of the pan of water at five-minute intervals as it cools.

An architect will check your results.

Staying cool for less is what you should be doing.

An air conditioner's cost depends on how much it runs.

The average of the control temperatures should be the same in all your tests.

There are a lot of important applications for second-order linear differential equations.

The equivalent IVP will be accepted by ODE Architect.

The periodic oscillation will continue indefinitely according to the equation.

2 are constants.

Figure 4.3 shows a graphical example of a solution.

The table needs to be polished and the mass streamlined so that it doesn't have any damping.

It is small but not zero.

There are 2 different amplitudes.

There are two types of seismographs.

Pendulums that respond to the motion of the ground are used in both of these instruments.

The axis is horizontal.

The ODE is used in Screen 2.3.

For more information on the architect, see Module 1 and Chapter 1

Ski jumpers take off from a ski jump.

We use boldface letters.

In a real-life situation, let's see how to use vectors.

The laws of motion can be expressed using vectors.

A professor is sitting over a dunk tank.

The target is 20 feet away from the ground.

The acceleration is due to gravity.

There is 0 in radians.

The answer is suggested by a mathematical model and ODE Architect.

His timing needs to be perfect.

Ski jumpers are subject to lift and drag forces.

Ski jumping is fun because of the lift force.

Positive constants are 0

Einstein was dunked by launch angles and speeds.

ODE Architect is playing a game.

A ball is thrown in the air.

ODE Architect can be used to determine whether the ball rises or falls.

IndianaNewton lands on the boxcar.

IndianaNewton lands on the boxcar.

When it comes to drag,Newton hits the train sooner.

IndianaNewton floats down.

The ski jump needs to be tilt upward.

There are physical problems that have two or more dependent variables.

ODEs can be changed into systems of first-order ODEs.

A model system of second-order ODEs is presented in module 6.

This is a good place to introduce matrix notation because it saves space and it appears as bold letters.

The formulas for this case are given in the texts listed in the references.

2 are constants.

We can see solutions graphically.

The corresponding points form a curve.

2 are not real constants.

The solution can be written in terms of real-valued functions.

There are 2 constants.

There is a graphing technique that is useful.

The behavior of the trajectory has to change as the changes occur.

Sometimes business enterprises are affected by seasonal influences.

In the case of Diffey and Cue, we can show this.

The periodic phenomena are often modeled with the help of the sine and cosine functions.

Interval is used to make clear the ultimate behavior of the solution.

Consider the tanks with salt water.

There is a diagram showing the tank system.

Limit the concentrations of salt in each tank.

Consider the tanks with salt water.

Draw a picture of the tank system.

ODE Architect can help you check your intuition.

Complex eigenvalues first occur in 1 and 4.

Motions are confined to a vertical plane by attaching all of them together.

We'll assume the system doesn't have a lot of damping.

There is a sketch of the double pendulum system.

There are twenty-four ways to spin a book.

It is possible to modify the ideas used to understand lin ear systems.

There are new phenomena that occur only in nonlinear systems.

When the phase portraits are superimposed, you can see the approximation.

Chapter 6 contains the negative real part of both eigenvalues.

The components are pretty much the same.

The following example shows that all bets are off.

A linear center is converted to a nonlinear sink in Figure 7.3.

The separatrices may change if we add some higher-order terms.

There are saddle separatrices along the axes and two of them are bent to the right.

The separatrices are bends by the nonlinearity.

Linear approximations near equilibrium points are used to investigate the long-term behavior of solution curves.

When we change the value of this parameter, the current, voltages and time circuit will change.

Use ODE Architect's equilibrium, Jacobian, and eigenvalue capabilities to check your work.

Let's say that a book has a uniform density.

It's obvious that 2 is obvious.

They are given in the margin sketch, which is more difficult to discern.

Put a tennis racket in the air and watch it move.

Let's not pay attention to air resistance.

It is another matter.

The body axis leads to strange movements.

There is a saddle point on the ellipsoid.

We won't pay much attention to the effects of damping in this system.

These equations are similar to the ones obtained before.

The term is a measure of the interaction between the two species.

To account for this, modify the predator-prey model.

How can it be made more or less effective?

Predicting the long-term behavior of the populations is possible with the model.

The spider-fly model is modified so that there are two competing to eat the flies.

The "Biological Models" folder contains the library file "A Predator-Prey System with Resource Limitation".

You can give a physical interpretation of your answers.

The bodies are spinning.

Explain what you see.

The coupled springs model has ODEs in Module 6.

The restoring force is 3.

Some 3D systems seem chaotic.

This is thought to be a feature of chaotic dynamics.

If you see an example of this type of chaos, change parameters.

A salt solution is pumped.

There is a single source of pollution that affects a well-defined habitat.

There are many medications that contain an antihistamine and a decongestant in a single capsule.

The architect is trying to track the flow of medication through the body.

6 units are 5 (hr)

Lead was used in food, air, and water, as well as lead-based paint, glaze, and crystalware.

It goes back into the blood from the two body compartments.

The allergy relief system is a cascade.

There are some references to the lead problem.

There is a step in the middle of the first-order reactions.

The behavior of the state suddenly changes when a bifurcation value is transited.

Chapter 7 is here.

It's not always easy to check the conditions for a Hopf bifurca tion.

To duplicate Figure 8.5, use ODE Archi tect.

The system's behavior carries over to the autocatalator.

The time interval is 100 with 1000 points.

The autocatalator's behavior will be similar to that of the system.

See the references for more on bifurcations.

If you take a pill every four hours, you'll get rid of the symptoms of a cold.

The following points are addressed in a report.

Imagine being an experimental pharmacologist for Get Well Phar maceuticals.

Lower and upper bounds will assure both effectiveness and safety.

There are schematic for chemical reactions.

Write out the ODEs for the individual concentrations.

These are Hopf cycles.

The architect is going to scroll through the trajectory.

Population biology studies how communities of organisms change.

Population biology includes ecol ogy, demography, population genetics, and epidemiology.

We'll look at mathematical models in epidemiology and ecology.

Modern society is seeing an increase in the awareness of environmental issues.

We want to keep the mathematical model simple, so we won't describe the biology or ecology of our population in detail.

Positive constants are 0

There are a few important features of our model.

Many forms can be taken in these interactions.

Simple models are the first step in modeling complicated ecologies.

The model gives insight into a number of important ecological issues, such as the nature of coexistence of predator and prey and the understanding of population cycles.

There is a model of two competing species.

Understanding how infectious diseases spread through populations is an important use of mathematical models.

Epidemiology is a field.

As individuals become susceptible, the population size changes.

Think of this process as converting susceptibles to infections.

conversion of susceptibles increases the population and decreases it when individuals recover.

There is a plot of susceptibles, infections, re covereds, and their sum.

The ODE Architect Library is open.

The solution curves are shown in the graphs.

You produced graphs for Problems 1 and 2.

Human societies use resources.

You will analyze a model for harvesting sin gle species.

The habitat is a forest and the resource is a pine tree.

The forest will become overexploited.

In natural populations, predator-prey interactions are very common.

Ecologists call such interactions mutualistic or symbiotic.

A simple model for mutualism will be presented in this exploration.

The competition model studied in Mod ule 9 is very similar to our model.

The equations that model a mutualistic interaction are in this file.

The SIR model is used for the spread of epidemics.

The equations have a time interval of 24 units.

The ODE is a model of the motion of a pendulum.

The solution curves of a pendulum system.

The ODEs have advantages that their solutions have explicit formulas.

Understand the differences by exploring all the options.

It's difficult to get explicit solution formulas because the integral can't be expressed in terms of elementary functions.

0 are ellipses.

1 gives rigidity.

The question is subtle.

This description will be turned into a differential equation.

It seems like it reflects our qualitative description.

Is always 1.

We can now focus on the torus.

From -8 to 8.

Arnold's book is the classical text.

Since the usual swingpumping is done without reference to a clock, a proper model must give an equation.

There is more on how to conserve energy in 6.5 and 8.1.

Screen 1.2 will help here.

Understand the motions of the pendulum.

An old spring is stretching.

The aging springs are modeled by a spring-mass equation.

This is stated more precisely by the following theorem.

There are a few things to keep in mind when looking for a Frobenius series.

The equation of order zero can be turned into a module.

Without ever going below zero.

The step function is used for damping.

At which time a constant force is applied.

Bessel functions are similar to decaying sinusoids.

Screen 2.6 is in Module 11.

Use ODE Architect to solve ODE.

A map of a pendulum superimposed on a trajectory.

The solutions of a forced damped pendulum ODE will be looked at in this chapter.

The study is more complicated now.

Chaos is needed in order to gain control.

This is similar to taking pictures with a strobe light.

There are saddles.

There are basins of attraction of the stable downward equilibrium posi tions of the unforced damped pendulum.

Near an unstable equilibrium, the behavior is more interesting.

The unforced pendulum was described above the Poincar'e plane.

There are basins for a forced damped pendulum.

It will do something else.

The tech niques used in the proof were originally developed by Smale4.

A mathematician named Stephen Smale was awarded a Fields medal in the 1960's.

We have shown how to control the pendu lum.

Chaos can be an undesirable aspect of physical motions.

There are two examples.

Let's go back to the general system.

A periodic point is a fixed point.

Nested functions can exhibit chaotic behavior as well as periodic behavior.

Transient behavior is eliminated by this.

Go to the One-dimensional tab.

Calculating a proper definition is still being argued about.

Due to Devaney, we'll use one to get the idea across.

There are examples in module 13

Chaotic dynamics can be found in a wide range of models.

The usual properties of exponential functions hold in the complex plane.

The iteration of a real function is the same as the iteration of a complex function.

Let's look at a simple example.

This leads to another question.

The tool is different.

The Mandelbrot set contains more information than is written here.

Eachblob of the set corresponds to a particular period.

blobs attract periodic behaviors

"fast" circles tend to be long.

The color scheme is up to the programmers.

You can play with a third system that uses complex numbers.

Maps is relevant to this chapter.

Three sources on chaos that are less technical and more accessible give an accurate description of what chaos is all about.

A closed,Bounded and Discon Chaotic behavior never repeats, and every set has uncountably many points.

In Mod left, there is an equivalent list of requirements.

A widely used model for population dynamics.

A set with no islands is a connected set.

If you were to choose the same initial data, you could choose any system of ODEs.

The number of times a cycle is repeated is its period.

Limit cycle is used for a cycle in a system of 2 ODEs.

If we are dealing with a grid of points.

The "kinetic energy" is 2.

Module 13 can be seen.

A fixed point is a part of the system.

2 are constants.

It is an equilibrium.

You can use a ball and string to test this.

The choice comes to an end.

The growth rate can be reduced by Robert Hooke, an English physicist.

See the initial value problem.

Linear algebra can be used to find the general solu A cycle.

0 does not go to infinite.

The first picture was published in 1978.

The equations need to be solved.

To see if the solution is reasonable.

There are two normal fre quincies.

You can see Module 12 Screen 1.4.

Any reaction can be extended with these ideas.

The equilibrium is the source of Separatrices.

There is a Hooke's law restoring force.

A directed quantity is a vector.

Document Outline

Cover

About the Authors

Table of Contents Ch 1 Introduction 1.1 Some Basic Mathematical Models; Direction Fields 1.1 Problems 1.2 Solutions of Some Differential Equations 1.2 Problems 1.3 Classification of Differential Equations 1.3 Problems 1.4 Historical Remarks Ch 2 First Order Differential Equations 2.1 Linear Equations with Variable Coefficients 2.1 Problems 2.2 Separable Equations 2.2 Problems 2.3 Modeling with First Order Equations 2.3 Problems 2.4 Differences Between Linear and Nonlinear Equations 2.4 Problems 2.5 Autonomous Equations and Population Dynamics 2.5 Problems 2.6 Exact Equations and Integrating Factors 2.6 Problems 2.7 Numerical Approximations: Euler's Method 2.7 Problems 2.8 The Existenc7&YgkYEN*o>>+7OrPS,AuV5 2.8 Problems 2.9 First Order Difference Equations 2.9 Problems Ch 3 Second Order Linear Equations 3.1 Homogeneous Equations with Constant Coefficients 3.1 Problems 3.2 Fundamental Solutions of Linear Homogeneous Equations 3.2 Problems 3.3 Linear Independence and the Wronskian 3.3 Problems 3.4 Complex Roots of the Characteristic Equation 3.4 Problems 3.5 Repeated Roots; Reduction of Order 3.5 Problems 3.6 Nonhomogeneous Equations; Method of Undetermined Coefficients 3.6 Problems 3.7 Variation of Parameters 3.7 Problems 3.8 Mechanical and Electrical Vibrations 3.8 Problems 3.9 Forced Vibrations 3.9 Problems Ch 4 Higher Order Linear Equations 4.1 General Theory of nth Order Linear Equations 4.1 Problems 4.2 Homogeneous Equations with Constant Coeffients 4.2 Problems 4.3 The Method of Undetermined Coefficients 4.3 Problems 4.4 The Method of Variation of Parameters 4.4 Problems Ch 5 Series Solutions of Second Order Linear Equations 5.1 Review of Power Series 5.1 Problems 5.2 Series Solutions near an Ordinary Point, Part I 5.2 Problems 5.3 Series Solutions near an Ordinary Point, Part II 5.3 Problems 5.4 Regular Singular Points 5.4 Problems 5.5 Euler Equations 5.5 Problems 5.6 Series Solutions near a Regular Singular Point, Part I 5.6 Problems 5.7 Series Solutions near a Regular Singular Point, Part II 5.7 Problems 5.8 Bessel's Equation 5.8 Problems Ch 6 The Laplace Transform 6.1 Definition of the Laplace Transform 6.1 Problems 6.2 Solution of Initial Value Problems 6.2 Problems 6.3 Step Functions 6.3 Problems 6.4 Differential Equations with Discontinuous Forcing Functions 6.4 Problems 6.5 Impulse Functions 6.5 Problems 6.6 The Convolution Integral 6.6 Problems Ch 7 Systems of First Order Linear Equations 7.1 Introduction 7.1 Problems 7.2 Review of Matrices 7.2 Problems 7.3 Systems of Linear Algebraic Equations; Linear Independence, Eigenvalues, Eigenvectors 7.3 Problems 7.4 Basic Theory of Systems of First Order Linear Equations 7.4 Problems 7.5 Homogeneous Linear Systems with Constant Coefficients 7.5 Problems 7.6 Complex Eigenvalues 7.6 Problems 7.7 Fundamental Matrices 7.7 Problems 7.8 Repeated Eigenvalues 7.8 Problems 7.9 Nonhomogeneous Linear Systems 7.9 Problems Ch 8 Numerical Methods 8.1 The Euler or Tangent Line Method 8.1 Problems 8.2 Improvements on the Euler Method 8.2 Problems 8.3 The Runge-KuttaUZ5f=* 8.3 Problems 8.4 Multistep Methods 8.4 Problems 8.5 More on Errors; Stability 8.5 Problems 8.6 Systems of First Order Equations 8.6 Problems Ch 9 Nonlinear Differential Equations and Stability 9.1 The Phase Plane; Linear Systems 9.1 Problems 9.2 Autonomous Systems and Stability 9.2 Problems 9.3 Almost Linear Systems 9.3 Problems 9.4 Competing Species 9.4 Problems 9.5 Predator-Prey Equations 9.5 Problems 9.6 Liapunov's Second Method 9nj~>b0?3Q 9.7 Periodic Solutions and Limit Cycles 9.7 Problems 9.8 Chaos and Strange Attractors; the Lorenz EquL1AE] 9.8 Problems Ch 10 Partial Differential Equations and Fourier Series 10.1 Two-Point Boundary Valve Problems 10.1 Problems 10.2 Fourier Series 10.2 Problems 10.3 The Fourier Convergence Theorem 10.3 Problems 10.4 Even and Odd Functions 10.4 Problems 10.5 Separation of Variables; Heat Conduction in a Rod 10.5 Problems 10.6 Other Heat Conduction Problems 10.6 Problems 10.7 The Wave Equation; Vibrations of an Elastic String 10.7 Problems 10.8 Laplace's Equation 10.8 Problems Appendix A. Derivation of the Heat Conduction Equation Appendix B. Derivation of the Wave Equation Ch 11!|U"87"AzM* 1/4 (tm)o7WF`B]}EMa1Feezy3C{ooAS3@*a/ki= 11.1 The Occurrence of Two Point Boundary Value Problems 11.1 Problems 11.2 Sturm-Liouville Boundary Value Problems 11.2 Problems 11.3 Nonhomogeneous Boundary Value Problems 11.3 Problems 11.4 Singular Sturm-Liouville Problems 11.4 Problems 11.5 Further Remarks on the Method of Separation of Variables: A Bessel Series Expansion 11.5 Problems 11.6 Series of Orthogonal Functions: Mean Convergence 11.6 Problems

Index

Student Solutions Manual