Intro. to Differential Equations

[ExtravagantDreams]

My intent is to create a thread for people interested in Differential Equations. However, I will explicitly state that I am only a student of this class myself and that many things could end up being incorrect or an improper way to present the material.

I will merely be going along with the class, mostly using excerpts and questions from the book, "Elementary Differential Equations and Boundary Value Problems: Seventh Edition," by William E. Boyce and Richard C. DiPrima. So truthfully, this is more for myself. Looking things up and explaining it to others seems to be the best way to learn.

If people have any questions or comments, feel free to share. Also, I know there are many knowledgeable people on this board, so be sure to correct me or make suggestions.

This will require knowledge of Calculus but don't be shy to ask if there is something that you are unsure of.


First, a little background;

What is a Differential Equation?
A Differential Equation is simply an equation containing a derivative.

Classifications:

Ordinary Differential Equations (ODE) - Equations that appear with ordinary derivatives (single independent variable, could have multiple dependent variables).
Examples:


\frac {dy} {dt} = ay - b



a \frac {dy_1} {dx} + b \frac {dy_2} {dx} + cy_1 = dy_2 = e


Partial Differential Equations (PDE) - Equations that appear with partial derivatives (multiple independent variable).
Examples:


\alpha^2 [ \frac {\partial^2 u(x,t)} {\partialx^2} ] = \frac {\partialu(x,t)} {\partialt}



\frac {\partial^2 V(x,y)} {\partialx^2} + \frac {\partial^2 V(x,y)} {\partialy^2} = 0



Don't let any of this frighten you. Math is always scary when looked at a glance with a bunch of undefined variables.

Linear and Nonlinear
The ordinary differential equation:

F(t, y, y', ..., y^{(n)}) = 0

is said to be linear if F is a linear function of the variables y, y',..., yn (Dependant variable must be first order). Thus the general linear ordinary differential equation of order n is:

a_0 (t) y^{(n)} + a_1 (t) y^{(n-1)} + ... + a_n (t) y = g(t)


where (n) is not the power of but the nth derivative.

An example of a simple Nonlinear ODE would simply be:

y \frac {dy} {dx} = x^4

This concludes the introduction. I may or may not write the next chapter tonight. However, a question, does anyone know an easier way for writing math on the computer and one that looks less confusing. I know I will have difficulty finding some things, especially subscripts and superscripts. Anyone know a better way to denote these?

[Greg Bernhardt]

Sounds great! Tutorials like this have been very successful here.

Howto make math symbols:
http://physicsforums.com/announcement.php?forumid=73

You can make subscripts and subscripts by using these tags

[ sup ] content [ /sup ]
[ sub ] content [ /sub ]

* no spaces

[ExtravagantDreams]

"This chapter deals with differential equations of the first order


\frac {dy} {dt} = f(t,y)


where f is a given function of two variables. Any differentiable function y = Φ(t) that satisfies this equation for all t in some interval is called a solution."

Linear Equations with Variable Coefficients

Using the previous example for ODE (dy/dx = ay + b) and replacing the constants we write the more general form:


\frac {dy} {dt} + p(t) y = g(t)

or

y' + p(t) y + g(t) = 0


where p(t) and g(t) are given functions of the independent variable t.

Special cases:
If p(t) = 0 then,


y' = g(t)


and the integral is easily taken;


\frac {dy} {dt} = g(t)



\int (\frac {dy} {dt}) dt = \int g(t) dt



y = \int g(t) dt + C


If g(t) = 0, then,


y' = p(t) y


and the integral is once more relatively easy to take;


\frac {dy} {dt}= p(t) y



\int \frac {dy} {y} = \int p(t) dt



ln|y| = \int [p(t) dt] + C



e^{ln|y|} = e^{\int [p(t) dt] + C}



y = Ke^{\int [p(t) dt] + C}, K = ± e^{C}


However, if neither p(t) or g(t) are zero in the general equation, a function µ(t) (the integrating factor) is used to solve the equation;


\mu (t) \frac {dy} {dt} + \mu (t) p(t) y = \mu (t) g(t)


where now the left hand side will be a known derivative


\mu (t) \frac {dy} {dt} + \mu (t) p(t) y = \frac {d} {dt}[\mu (t)y]


so that, in theory, you end up with;


\frac {d} {dt}[\mu (t)y] = \mu (t) g(t)


Since µ(t) must be carefully chosen to make the previous statement true, let us find it.


\frac {d} {dt}[\mu (t)y] = \mu (t) \frac {dy} {dt} + \mu (t)p(t)y



\frac {d} {dt}[\mu (t)y] = \mu (t) \frac {dy} {dt} + \frac {d \mu (t)} {dt} y


Where the latter is simply the derivative of µ(t)y in general form.


\mu (t) \frac {dy} {dt} + \mu (t) p(t) y = \mu (t) \frac {dy} {dt} + \frac {d\mu (t)} {dt} y


Subtracting µ(t)(dy/dt) from both sides


\frac {d \mu (t)} {dt} y = \mu (t) p(t) y


Cancel y


\frac {d \mu (t)} {dt} = \mu (t) p(t)



\frac {d \mu (t)} {\mu (t)} = p(t) dt



\int \frac {d \mu (t)} {\mu (t)} = \int p(t)d t



ln|\mu (t)| = \int p(t) dt


The constant C is arbitrary and can be dropped to form the equation,

µ(t) = e∫[p(t)dt]

So the integrating factor µ(t) can always be found by the last equation.

Lets try some problems together.

Ex1.
y' + 2y = 3

In this equation p(t) = 2 and g(t) = 3. Since neither of them are zero, use an the integrating factor µ(t) to create a differentiable equation,

µ(t)y' + µ(t)2y = µ(t)3

Solve µ(t)to be,

µ(t) = e∫[p(t)dt]
µ(t) = e∫[2dt]
µ(t) = e2t

Plug the value of µ(t) back into the equation to obtain,

e2ty' + e2t2y = e2t3

Recognize that the left hand side of the equation is merely [ye2t]',

[ye2t]' = d/dt[ye2t] = 3e2t
∫d/dt[ye2t] = ∫3e2t
ye2t = (3/2)e2t + C
y = (3/2) + Ce-2t

Ex2.
y' + (1/2)y = 2 + t
µ(t)y' + (1/2)µ(t)y = (2 + t)µ(t)

µ(t) = e∫[p(t)dt] = et/2

et/2y' + (1/2)et/2y = 2et/2 + tet/2
d/dt[et/2y] = 2et/2 + tet/2
et/2y = ∫[2et/2 + tet/2]dt
et/2y = 4et/2 + ∫[tet/2]dt

Using integration by parts,

u = t, du = dt
v = 2et/2, dv = et/2dt

∫[tet/2]dt = 2tet/2 - ∫[2et/2]dt
∫[tet/2]dt = 2tet/2 - 4et/2 + C

et/2y = 4et/2 + 2tet/2 - 4et/2 = 2tet/2 + C
y = 2t + Ce-t/2

For initial value problems it is easy to solve for C. Taking the last problem, solve if
y(0) = 2

2 = 2(0) + Ce-(0)/2 = C
C = 2

Therefore,
y = 2t + 2e-t/2


That is enough for now. Here are some problems to practice on if you so wish.

1.) y' + 3y = t + e-2t
2.) 2y' + y = 3t, hint: rewrite to fit general equation y' + p(t)y = g(t)
3.)t3(dy/dt) + 4t2y = e-t, y(-1) = 0

[ExtravagantDreams]

Thanks Greg, I will run through it tomorrow and change it to make it more readable.

[Integral]

Well I had to dig out my copy of Boyce and Diprima (2nd Edition!) to follow your development, it all works out as you have presented.

I will follow along with you, relearning what I have not see for a number of years, and perhaps able to help you out if you hit some rough spots.

[Integral]

Here is solution (http://home.comcast.net/~Integral50/Math/diffeq1.PDF) to the first exercise.

[ExtravagantDreams]

Thank you for your participation. Your solution is infact correct except for the constant is missing. No biggy, I always forget those too. Did you find it hard to follow without the book and should I have presented this more clearly some how?

I'm glad to know that someone else knows this stuff. I have some trouble understanding finding the interval for nonlinear functions for which a solution exists, so if I havn't figured it out by the time I do the write up, perhaps you can help.

[Integral]

It was very sloppy of me to leave off the constant, sorry about that.

I was a bit confused by your presentation as it is light on connective text. Where you presented

\frac {d} {dt}[\mu (t)y] = \mu (t) g(t)
I was thrown for a bit. My copy of B&D helped out. The fact is everything you wrote is absolutely correct.

I have taken grad level ODE & PDE courses in the dim and distant past ('86-'88 time frame) So should be able to dredge up some long buried knowledge to help out.

I have always found Differential Equations to be interesting, you might say they are where math and reality meet. With a good back ground in Diff Eqs and some numerical methods you can do dang near anything.

edit: corrected symbols

[Integral]

Here is the solution (http://home.comcast.net/~Integral50/Math/diffeq3.PDF) to the 3rd exercise.

[ExtravagantDreams]

Yes, your answer is correct. It actually took me a while to get it. Out of curiousity, why did you change to using the variable s? Are you just used to using it and forgot that it was in terms of t or can this be done?

[Integral]

In the integral

\int \mu (s) g(s)ds
The variable, s, is what is called a dummy variable, it can be anything. You will see this frequenty.

[StephenPrivitera]

Integral, how come your solutions don't show up?

[Integral]

They are PDFs do you have acrobat reader installed?

[StephenPrivitera]

Yeah I do. I didn't realize your text was a link.

[StephenPrivitera]

Integral, for the last part of your solution to the thrid exercise, I get
y(t)=t-4 [inte]te-tdt=t-4[-te-t-e-t + C]
So the t-4 is distributed through the C.
y(t)=-t-4e-t(t + 1)+ t-4C
Applying the initial condition, I get C=0, so
y(t)=-t-4e-t(t + 1). We get the same result, but through different ways.

[ExtravagantDreams]

I'm terribly sorry, I have been entirely too busy recently but I'm back, for the moment.

So now that we know how to differentiate a first order linear equation with variable coefficients, let us move on to linear separable equations.

In the last section we used the form dy/dt = ay + b, where the more general form of first order equation is;

dy/dx = f(x,y)

If this function can be rewritten in the form
M(x) + N(y)dy/dx = 0

where M is a function of x only and N is a function of y only, then the function is said to be separable. It can also be written in the form of
M(x)dx + N(y)dy = 0

It is as simple as that.

Let's try a couple examples.

Ex. 1
dy/dx = x2/(1 - y2)

-x2 + (1 - y2)dy/dx = 0
∫-x2dx + ∫[(1 - y2)(dy/dx)]dx = ∫0dx
-x3/3 + (y - (y3)/3) = C

Where the answers can be left in a few different forms;
-x3/3 + (y - (y3)/3) = C
-x3 + 3y - y3 = 3C = C1
3y - y3 = x3 + C1


The "cheating" way would be to cross multiply the initial equation. Although, this is not correct, it will end in the same answer.

dy/dx = x2/(1 - y2)
∫(1 - y2)dy = ∫x2dx
y - (y3)/3 = x3/3 + C
3y - (y3) = x3 + C1


Let's try one with an initial condition

Ex. 2
dy/dt = ycost/(1 + 2y2), y(0) = 1

(1 + 2y2)dy/dt = ycost
-cost + ((1 + 2y2)/y)dy/dt = 0
∫-cos(t)dt + ∫[((1 + 2y2)/y)dy/dt]dt = ∫0dx
-sint + ∫[y-1 + 2y]dy = C
-sint + ln|y| + y2 = C
ln|y| + y2 = sint + C

ln|1| + 12 = sin(0) + C
C = 1

ln|y| y2 = sint + 1

Again, this differential equation can be solved the "cheating" way by cross multiplying,

dy/dt = ycost/(1 + 2y2), y(0) = 1

[(1 + 2y2)/y]dy = cos(t)dt
∫(y-1 + 2y)dy = ∫cos(t)dt
ln|y| y2 = sint + C


If you so desire, here are some problems

1. y' = x2/y
2. xdx + ye-xdy = 0, y(0) = 1
3. y2(1 + x2)1/2 + arcsinx dx

[StephenPrivitera]

The "cheating" way would be to cross multiply the initial equation. Although, this is not correct, it will end in the same answer.
____________
I've often wondered why this is the case. dy and dx are variables right? So why can't they be treated as such?

[StephenPrivitera]

Here's my solution to the first one:
dy/dx=x2/y
y(dy/dx)=x2
[inte] y(dy/dx)dx = [inte] x2dx
(1/2)y2= (1/3)x3 + C1
y2= (2/3)x3 + C2

[ExtravagantDreams]

"I've often wondered why this is the case. dy and dx are variables right? So why can't they be treated as such?"

x is a variable, y is a funtion of x. By cross multiplying you would be treating y like a variable.

And your answer is correct.

[ExtravagantDreams]

I will mention right now, I have some difficulty understanding this part so I will have even more difficulty explaining it and will probably need a little help. The good news, this isn't the most important topic in my opinion.

The reason this is applicable is that it is often easier to find the existence and uniquness of a solution without having to work out the actual problem.

Theorm:
Let the function f and &part;f/&part;y be continuous in some rectangle [alpha] < t < [beta], &Gamma; < y < &delta; containing the point (t0, y0). Then, in some interval t0 - h < t < t0 + h contained in [alpha] < t < [beta], there is a unique solution y = &phi;(t) of the initial value problem
y' = f(t,y), y(t0) = y0


Assuming that both the function and it's partial deivative with respect to y are continuous in a rectangle R; |t - t0| < A, |y - y0| < B;

Let:
M = max |f(t,y)|, (t,y) is in R
C = min (A,B/M)

Then:
There is one and only one solution y(t) valid for
t0- C < x < t0+ C

http://www.angelfire.com/tx5/extravagantdreams/Nonlinear_Interval.jpg

Ex.
y' = y2, y(0) = 1
f(t,y) = y2
&part;f/&part;y = 2y

M = max |f(t,y)|, (t,y) is in R
C = min (A,B/M)

Since A can be made infinately large find the max for B

M = max|(1 + B)2|
C = B/M

C = B/(1 + B)2
C = 1/4

If you have no idea where the answer came from, you are in the same boat as I. If you do know, please share.

Perhaps another example

Ex. 2
y' = (t2 +y2)3/2, y(t0) = y0
&part;f/&part;y = 3y(t2 + y2)1/2

M = max|(t2 +y2)3/2|
M = [(t0 + A)2 + (y0 + B)2]3/2

C = min(A,B/M)
C = min(A, B/[(t0 + A)2 + (y0 + B)2]3/2)


Also, does anyone know how to display determinants or matracies on the computer?

[Integral]

Keep an eye on what you are doing in the above examples. You are finding a specific region upon which you can guarantee that a unique solution exists. This means you must provide some specific numbers to the bounds of the region.

One thing that might help you a little, redraw your picture with the A and B intervals CENTERED on (t0,y0)

In your worked example the interval for y is:
B-y0<=y <= B+y0

Since y0=1 we have:
1-B<=y<=B+1

so if M= Max|f(t,y)| and f(t,y)= y2 The maximum value of f(t,y) on our interval is (B+1)2
Since A can be made infinitely large find the max for B

I believe that this is saying you are free to pick any value on the t axis for t0, once that value is defined, the value of f(t[0,y) will be fixed and you can guarantee the existence of a Max value for f(t,y)On an interval surrounding it.C = B/(1 + B)2
This follows from the definitions.
C = 1/4
To get this a value of B must be given or defined, since you are free to choose B to be anything, it looks like someone picked 1.

Does this help?

[ExtravagantDreams]

I guess I am getting confused because the book does not use
M = max |f(t,y)|
C = min (A,B/M)

notation. From what I understand, all this really says is that you are trying to find the maximum interval of f(t,y), where the limiting factor is either A or B/M (which is denoted C). But why B/M, why not just B? This is what gets me. Is C a value in the horizontal direction? Does dividing B by M make it a horizontal value? And how do you choose B?

I think I am making this too difficult. I fear it is one of those things where you just have to look at it.

[ExtravagantDreams]

Theorem:
Let the functions M, N, My, Nx, where subscripts denote partial derivatives, be continuous in the rectangular region R: [alpha] < [beta], &Gamma; < y < &delta;. Then;

M(x,y) + N(x,y)y' = 0

is an exact differential equation in R if and only if

My(x,y) = Nx(x,y)

at each point of R. That is, there exists a function satisfying

&part;[psi]/&part;x(x,y) = M(x,y), &part;[psi]/&part;y(x,y) = N(x,y)
or
[psi]x(x,y) = M(x,y), [psi]y(x,y) = N(x,y)

if and only if M and N satisfy

My(x,y) = Nx(x,y)


This means that there is a solution [psi](x,y) of the general equation M + Ny' = 0

Proof of My(x,y) = Nx(x,y)
We already defined

[psi]x(x,y) = M(x,y), [psi]y(x,y) = N(x,y)

we can compute the partial derivative of each to be

[psi]xy(x,y) = My(x,y), [psi]yx(x,y) = Nx(x,y)

Since My and Nx are continuous, it follows that [psi]xy and [psi]yx are continuous also, which also guarantees their equality.

Finding [psi](x,y)
when My(x,y) = Nx(x,y):
Starting with the equations

[psi]x(x,y) = M(x,y), [psi]y(x,y) = N(x,y)

start with the first and integrate with respect to x

[psi](x,y) = &int;M(x,y)dx + h(y)

Where h is some function of y playing the role of an arbitrary constant. Now we must proove that h(y) can always be chosed so that [psi]y = N

[psi]y(x,y) = &part;/&part;y[&int;M(x,y)dx + h(y)]
[psi]y(x,y) = &int;My(x,y)dx + h'(y)

Setting [psi]y = N we obtain

N(x,y) = &int;My(x,y)dx + h'(y)

Where we can then solve for h'(y)

h'(y) = N(x,y) - &int;My(x,y)dx
h(y) = Nx(x,y) - My(x,y)

Then the general solution

[psi](x,y) = &int;M(x,y)dx + &int;[N(x,y) - &int;My(x,y)dx]dy


Ex. 1

2xy3 + 3x2y2y' = 0

Where M = 2xy3, N = 3x2y2
Then My = 6xy2, Nx = 6xy2

Since My = Nx
This equations is exact and can be solved by the previous method.

Start with

[psi]x = M = 2xy3
[psi] = &int;Mdx + h(y) = &int;2xy3dx + h(y)
[psi] = x2y3 + h(y)

Find h;
Since we know that N = &part;[psi]/&part;y, differentiate both sides and substitude N

N = &part;/&part;y[&int;Mdx + h(y)
N = 3x2y2 + h'(y)

h'(y) = 0, h(y) = C

Plug h(y) back into the original [psi] equation
[psi] = x2y3 + C

[psi] = x2y3 = K

Ex. 2
Find the function [psi](x,y) of
y' = -(ax + by)/(bx - cy)

(bx - cy)y' = -(ax + by)
(ax + by) + (bx - cy)y' = 0

So
M = (ax + by), N = (bx - cy)
My = b, Nx = b

Since My = Nx, the equation is exact

[psi]x = M = ax + by
&int;[psi]xdx = &int;(ax + by)dx
[psi] = (1/2)ax2 + byx + h(y)
[psi]y = &part;/&part;y[(1/2)ax2 + byx + h(y)]
[psi]y = bx + h'(y)

[psi]y= N

N = bx + h'(Y)

(bx - cy) = bx + h'(y)

h'(y) = -cy
h(y) = -(1/2)cy2

[psi] = (1/2)ax2 + byx - (1/2)cy2 = K


Ex. 3
(ycosx + 2xey) + (sinx + x2ey - 1)y' = 0

So,
M = ycosx + 2xey, N = sinx + x2ey - 1
My = cosx + 2xey, Nx = cosx + 2xey

My = Nx, so can be solved using the exact method

Remember that
M = [psi]x
N = [psi]y

&int;[psi]xdx = &int;(ycosx + 2xey)dx
[psi] = ysinx + eyx2 + h(y)

Taking a partial derivative of both sides with respect to y,

[psi]y = sinx + x2ey + h'(y)

[psi]y = N = sinx + x2ey - 1

sinx + x2ey - 1 = sinx + x2ey + h'(y)
h'(y) = -1
&int;h'(y)dy = h(y) = -y

[psi](x,y) = ysinx + eyx2 - y = k


Integrating Factor
If an equation is not exact it can be multyplied by an integrating factor [mu] so that it becomes exact.

Starting with the general form
M(x,y) + N(x,y)y' = 0

and
My [x=] Nx

There is an integrating factor [mu] such that
[mu]M(x,y) + [mu]N(x,y)y' = 0

and
([mu]M)y = ([mu]N)x

or

[mu]yM + [mu]My = [mu]xN + [mu]Nx
M[mu]y - N[mu]x + (My - Nx)[mu] = 0

We will not discuss finding [mu](x,y) since this is entirely too diffictuly for this course, so we shall stick to [mu] as a function of x or y only.

If [mu] is a function of x;
([mu]M)y = [mu]My, ([mu]N)x = [mu]Nx + N(d[mu]/dx)

Thus, if ([mu]M)yis to equal [mu]Nx;
d[mu]\dx = (My - Nx)[mu]/N

If the function (My - Nx)/N depends on x only then [mu] can also be a function of x only and an integrating factor has been found.

The proceedure to finding [mu](y) is similar and the equation
d[mu]\dx = -(My = Nx)[mu]/M
is derived.

Ex. 4
(3xy + y2) + (x2 + xy)y' = 0
M = 3xy + y2, N = x2 + xy
My = 3x + 2y, Nx = 2x + y

Since,
My [x=] Nx

the equation is not seperable and an integrating factor must be found

[mu](x)
(My - Nx)/N dx = d[mu](x)/[mu](x)
ln|[mu](x)| = &int;(3x + 2y - 2x - y)/(x2 + xy)dx
ln|[mu](x)| = &int;(x + y)/x(x + y)dx
ln|[mu](x)| = &int;(1/x)dx
ln|[mu](x)| + ln|x|
[mu](x) = x

Since it is a function of x only, this can be used as the integrating factor. Now to solve the equations

x(3xy + y2) + x(x2 + xy)y' = 0
M = 3x2y + xy2, N = x3 + x2y
My = 3x2 + 2xy, Nx = 3x2 + 2yx

My = Nx
So it is now exact.

&part;/&part;y[&int;(3x2y + xy2)dx] = x3 + x2y
&part;/&part;y[yx3 + (1/2)y2x2 + h(y)] = x3 + x2y
x3 + x2y + h'(y) = x3 + x2y

h'(y) = 0, h(y) = C

yx3 + (1/2)y2x2 = K

Problems:

1. (2x + 3) + (2y - 2)y' = 0
2. (9x2 + y - 1)dx - (4y - x)dy = 0, y(1) = 0
3. (x2y = 2xy + y3)dx + (x2 + y2)dy = 0

[Integral]

Originally posted by ExtravagantDreams
I guess I am getting confused because the book does not use
M = max |f(t,y)|
C = min (A,B/M)

notation. From what I understand, all this really says is that you are trying to find the maximum interval of f(t,y), where the limiting factor is either A or B/M (which is denoted C). But why B/M, why not just B? This is what gets me. Is C a value in the horizontal direction? Does dividing B by M make it a horizontal value? And how do you choose B?

I think I am making this too difficult. I fear it is one of those things where you just have to look at it.

This is covered pretty well in my copy of B&D.

He is using Picard iterations to prove the existance of solutions. I'll cheat for this cause he says it better then I. Here is are links to scans of Boyce & Deprima 2nd Ed.First page (http://home.comcast.net/~rossgr1/Math/B_D1.jpg) second page (http://home.comcast.net/~rossgr1/Math/B_D2.jpg)
Hope this helps, is this not in your book?

[ExtravagantDreams]

A second order differential equation is linear if it is in the following form;
y'' + p(t)y' + q(t)y = g(t)
or
P(t)y'' + Q(t)y' + R(t)y = g(t)

Where p, q, and g and functions of t

If the problem has initial conditions it will be in the form of
y(t) = y0, y'(t) = y'0.

Let's start with the easiest:
Homogeneous with Constant coefficients

Homogeneous in this case means g(t) = 0,
with constant coefficients mean P(t) = p, Q(t) = q, and R(t) = r
Since makes the general equations;

py'' + qy' + ry = 0

Solving the homogeneous equation will later always provide a way to solve the corresponding nonhomogeneous problem.

I'm not going to proove all this but you can take the kernal of this funtion as

ar2 + br + c = 0

and you can, so to speak, find the roots of this funtion.

r1,2 = (-b &plusmn; &radic;(b2 -4ac))/2a

r1 = (-b + &radic;(b2 -4ac))/2a
r2 = (-b - &radic;(b2 -4ac))/2a

Assuming that these roots are real and different then;
y1(t) = er1t
y2(t) = er2t

and y = C1y1(t) + C2y2(t)

Which comes from the initial deriviation. If someone really really wants to know, I will show you.

y = C1er1t + C2er2t

This is your general solution.
C1 and C2 can be solved for if initial conditions y(t) and y'(t) are given in the following manner;

y = C1er1t + C2er2t

y(t) = y0

y0 = C1er1t + C2er2t

and

y' = r1C1er1t + r2C2er2t


y'(t) = y'0


It is also possible to varify your solution by using
y = C1er1t + C2er2t
y' = r1C1er1t + r2C2er2t
y'' = r12C1er1t + r22C2er2t

and pluging them back into the equation
ay'' + by' + cy = 0

Ex. 1
y'' + 5y' +6y = 0, y(0) = 2, y'(0) = 3

1r2 + 5r + 6 = 0

(r + 3)(r + 2)

r1 = -3
r2 = -2

y = C1e-3t + C2e-2t

2 = C1e-3(0) + C2e-2(0)
2 = C1 + C2


y' = -3C1e-3t - 2C2e-2t

3 = -3C1e-3(0) - 2C2e-2(0)
3 = -3C1 - 2C2

|+1 +1 +2|
|- 3 - 2 +3| ~

|+1 +1 +2|
|+0 +1 +9| ~

|+1 +0 - 7|
|+0 +1 +9| ~

C1 = -7
C2 = 9

y = -7e-3t + 9e-2t
y' = 21e-3t - 18e-2t
y'' = -63e-3t + 36e-2t

y'' + 5y' +6y = 0
(-63e-3t + 36e-2t) +5(21e-3t -18e-2t) =6(-7e-3t + 9e-2t)

e-3t(105 - 63 - 42) + e-2t(36 - 18 - 18) = 0

[hawaiidude]

ok heres question.... solve the differential equation d^3x/dt^3-2d^2d^2x/dt^2 +dx/dt=0



and find the fourier series for f(x)={0,pi -pi<x<0
0<x<pi

[ExtravagantDreams]

I'm actually not entirely sure how to solve this. This is not something we covered (higher order linear equations)

Skimming through the section, it looks like it is done in the same way as regular constant coefficient homogeneous equations.

3r3 - 2r2 + 1r = 0

where r are the kernals

r(3r2 - 2r + 1) = 0

r1 = 0
r23 = (1 &plusmn; i&radic;(2))/3


y = C1 + e1/3 x(C2cos((&radic;(2)/3)x) + C3sin((&radic;(2)/3)x))

and I'm not sure what a fourier series is.

[ExtravagantDreams]

I have fallen way behind in doing this, about 3 + 3 fairly complicated sections, so I will try to cover as much as I can during my time off tonight and tomorrow morning.

So, we already saw what happens to 2nd order linear homogeneous differential equations with constant coefficient (wow, that is a mouth full) in the form;

ay'' + by' + cy = 0

where one finds the roots by evaluating;

ar2 + br + c = 0

and the answer is in the form;

y = C1er1t + C2er2t

However, when the expression

b2 - 4ac

is negative, the roots can be written in the form;

r12 = &lambda; &plusmn; i&mu;

Where;
&lambda; = -b/2a
i&mu = &radic;(b2 - 4ac)
when b2 < 4ac

and

y1(t) = e(&lambda; + i&mu;)t
y2(t) = e(&lambda; - i&mu;)t

Let us look at some properties first:

Using taylor serious it can be prooven that
eit = cos(t) + isin(t)
e-it = cos(t) - isin(t)
ei&mu;t = cos(&mu;t) + isin(&mu;t)

Where then
e(&lambda; + i&mu;)t = e&lambda;tei&mu;t = e&lambda;t[cos(&mu;t) + isin(&mu;t)]

Looking for real solutions

Addition of function 1 and 2
y1(t) + y2(t) =
(e&lambda;t[cos(&mu;t) + isin(&mu;t)]) + (e&lambda;t[cos(&mu;t) - isin(&mu;t)]) = 2e&lambda;t[cos(&mu;t)]

Subtraction of function 1 and 2
y1(t) - y2(t) =
(e&lambda;t[cos(&mu;t) + isin(&mu;t)]) - (e&lambda;t[cos(&mu;t) - isin(&mu;t)]) = 2e&lambda;t[isin(&mu;t)]

Since the differential equation is made up of real coefficients, it can be said that it's derivative is also. Therefore by simply neglecting the constant multipliers we obtain a pair of real-valued solution;


u(t) = e&lambda;t[cos(&mu;t)]
v(t) = e&lambda;t[sin(&mu;t)]

where u and v are just the real and imaginary parts of the solution respectively; meaning these parts are linearly independent, and a combination of these parts is a then also a solution.

W(u,v)(t) [x=] 0

Since the Wronskian of these two functions is equal to &mu;e2&lambda;t, it is always nonzero as long as &mu; [x=] 0. Since &mu; is always greater then zero, when b2 < 4ac, the general solution is;

y = C1e&lambda;tcos(&mu;t) + C2e&lambda;tsin(&mu;t)

Ex.
y'' + y' + y = 0

r2 + r + 1

r12 = (-1 &plusmn; &radic;[(1)2 - 4(1)(1)])/2(1) =
-1/2 &plusmn; i&radic;(3)/2

Then;
&mu; = -1/2
&lambda; = &radic;(3)/2

y = C1e-t/2cos(&radic;(3)t/2) + C2e-t/2sin(&radic;(3)t/2)

[hawaiidude]

ok heres another one solve ......
:: .
x -6x+9x=y

how would u solve this

[ExtravagantDreams]

Originally posted by hawaiidude

x - 6x + 9x = y


LOL, are you trying to make fun of me?

What is the solution to the previous one?

[hawaiidude]

lol no it is correct

[Integral]

Originally posted by hawaiidude
ok heres another one solve ......
:: .
x -6x+9x=y

how would u solve this

Where are the differentials? You need to make this a differential equation. As it stands I get

y=4x

Not real exicting.

[ExtravagantDreams]

Originally posted by Integral
Where are the differentials? You need to make this a differential equation. As it stands I get

y=4x

Not real exicting.

Yeah, that's what I was thinking.

[ExtravagantDreams]

This is a method for finding a second solution to a 2nd order linear homogeneous differential equations with constant coefficients assuming you already have the first solution. It can often occur when the roots of the equation are the same (when b2 - 4ac = 0).

Lets remember that given the equation;

ay'' + by' + cy = 0,

where a, b, and c are constant coefficients,

There can be a solution found by first finding the roots of;

ar2 + br + c = 0

Then it is a simple matter of remember the equations;

C1er1t + C2er2t

for ordinary roots, and;

C1e&lambda;tcos(&mu;t) + C2e&lambda;tsin(&mu;t)

for complex roots in the form;

r = &lambda; ± i&mu;

The general idea is to find a non-linear multiple of the first solution. Given the first solution, y1 = ert, the general solution is C1ert and another solution y2 = v(t)y1, where v(t) is some funtion of t, can be found.

Using;
y2 = v(t)y1

Find;
y'2 = v'(t)y1 + v(t)y'1
y''2 = v''(t)y1 + 2v'(t)y'1 + v(t)y''1

And plug these into the original equation;

ay'' + by' + cy = 0

a[v''(t)y1 + 2v'(t)y'1 + v(t)y''1] + b[v'(t)y1 + v(t)y'1] + c[v(t)y1]

Now group the equation such that all v(t), v'(t), and v''(t) terms are together and if it was done correctly, all v(t) terms will cancel.

v''(t)[ay1] + v'(t)[2ay'1 + by1] + v(t)[ay''1 + by'1 + cv(t)y1]

Since,
b2 - 4ac = 0

ay''1 + by'1 + cv(t)y1 = 0

So you are left with;

v''(t)[ay1] + v'(t)[2ay'1 + by1]

Solve for v(t) by integrating and plug back into;
y2 = v(t)y1

Let's try an example:

y'' + 4y' + 4y = 0

Find the Roots;
r12 = [-4 &plusmn; &radic;(42 - 4(1)(4))]/2(1)
r12 = -2

y1 = e-2t

y2 = v(t)e-2t
y'2 = v'(t)e-2t - 2v(t)e-2t
y''2 = v''(t)e-2t - 4v'(t)e-2t + 4v(t)e-2t

Plug into original equation;
v''(t)e-2t - 4v'(t)e-2t + 4v(t)e-2t + 4[v'(t)e-2t - 2v(t)e-2t] + 4[v(t)e-2t]

Combine v(t) terms;
v''(t)[e-2t] + v'(t)[-4e-2t + 4e-2t] + v'(t)[4e-2t - 8e-2t + 4e-2t] = 0

v''(t)[e-2t] + v'(t)[ 0] + v'(t)[ 0] = 0

In this case the v'(t) terms also cancelled. So,

v''(t)[e-2t] = 0

Since,
e-2t [x=] 0,
v''(t) = 0

Integrating;
v'(t) = C1,
v(t) = C1t + C2,

Remember the original solution;
y1 = e-2t

and;
y2 = v(t)e-2t

y2 = [C1t + C2]e-2t
y2 = C1te-2t + C2e-2t

Where the second term is just the the first, so the general solution is;
y = C1te-2t + C2e-2t


It actually turns out that y2 = ty1, always.

Give these problems a try:

1. y'' - 2y' + y = 0
2. 4y'' + 12y' + 9y = 0

[hawaiidude]

hey thanks...nice examples,,,clear and easy to understand...but heres a problem...when x=0 3x^2y''-xy'+y=0

[ExtravagantDreams]

I'm not sure what you are asking;

3x2y''- xy'+ y = 0,

When x = 0

3(0)2y''- (0)y'+ y = 0

y = 0 ???

[hawaiidude]

not really but how do you find recurrence formulas? they're very complicated and i can't understand it...like the recurrence for (x^2+4)y''+xy=x+2

i thought you find the seond derivative and and first and the original

a0 )a1x+a2x^2+a3x^3+a4x^4.......+anx^n+an+1+an+2x^n+2 +...

y'=a1+2azx+3a3x^2...and so on..

how would you compute this? iam very confused ...all i got is you get the combining terms in this case8a2=2, 24a3+a0=1 2a2+48a4+a1=0 ...

then it's like n(n-1)an+4(n+2)(n+1)an+2+an-1=0 (n=0, 1 ,2 ,3 ,4 ....x

i know how they got the combined terms but how did they get the n's?

[ExtravagantDreams]

Are you talking about the sums in series solutsions of 2nd Order Linear Equations? If it is beyond that, sorry I can't help ya. I'm only taking this course just now.

[ExtravagantDreams]

Here we will look at Second Order, nonhomogeneous Linear Equations of the form;

L[y] = y'' + p(t)y' + q(t)y = g(t)

where p(t), q(t), and g(t) are continuous functions on an open interval, I. We can use the homogeneous equations, where g(t) = 0 to solve the nonhomogeneous.

If Y1 and Y2 are two solutions of the nonhomogeneous equation, then their difference (Y1 - Y2) is a solution of the corresponding homogeneous equation. If in addition, y1 and y2 are a fundamental set of solutions of the homogeneous equation, then;

Y1 - Y2 = c1y1(t) + c2y2(t)

where c1and c2 are certain constants.


The general solution of the nonhomogeneous equation can then be written as;

y = &phi;(t) = c1y1(t) + c2y2(t) + Y(t)

where y1 and y2 are a fundamental set of solutions of the corresponding homogeneous equation, c1 and c2 are arbitrary constants, and Y(t) is some specific solutions of the nonhomogeneous equation.

We will attempt here to find the function Y(t) of the nonhomogeneous equation. There are two methods of doing this, namely the method of undetermined coefficients, which is discussed in this section, and the method of variation of parameters, which will be discussed next time.

The idea is to assume a solution for Y(t) with an undetermined coefficient, then use this answer and plug back into the original equation and try to find the coefficient. If it is successfuly, a solution of Y(t) has been found, if not there is so solution in the form that was assumed and a differnt assumption should be made.

Clearly, this has draw back, such that an assumption must be fairly easy to do. Yet, once such an assumption has been made, the solution is not difficult to optain.

Let's look at some examples:

y'' - 3y' - 4y = 3e2t

Here we seek a function such that the combination of Y''(t) - 3Y'(t) - 4Y(t) = 3e2t

Since, the exponential function reproduces itself through differentiation it is the most plausible answer. Let's assume Y(t) = Ae2t, where A is the undetermined coefficient.

Y(t) = Ae2t
Y'(t) = 2Ae2t
Y''(t) = 4Ae2t

Plug these values into the combination equation;

1[4Ae2t] - 3[2Ae2t] - 4[Ae2t] = 3e2t

Attempt to solve for A;

[4 - 6 - 4]A = 3
A = -1/2

So;

Y(t) = -1/2e2t

Let us try another one, where we first assume the incorrect solution;

y'' - 3y' - 4y = 2sin(t)

Let us assume;

Y(t) = Asint(t)
Y'(t) = Acos(t)
Y''(t) = -Asint(t)

1[-Asint(t)] - 3[Acos(t)] - 4[Asint(t)] = 2sin(t)

A[-5 - 3cot(t)] = 2

Clearly, this can not be solved. Let's assume a differnt solution, namely Y(t) = Asint(t) + Bcos(t), where B is just another undetermined coefficient.

Y(t) = Asint(t) + Bcos(t)
Y'(t) = Acost(t) - Bsin(t)
Y(''t) = -Asint(t) - Bcos(t)

1[-Asint(t) - Bcos(t)] - 3[Acost(t) - Bsin(t)] - 4[Asint(t) + Bcos(t)] = 2sin(t)

[-A + 3B -4A]sin(t) + [-B - 3A - 4B]cos(t) = 2sin(t)

Since there there are two sin(t) on the right there must be two on the left, zero cos(t) on the right, there must be zero on the left.

Hence the coefficients of sin(t) and cos(t) must be;

-5A + 3B = 2,
-3A - 5B = 0

1 -3/5 -2/5
0 -34/5 -6/5

1 0 -5/17
0 1 3/17

A = -5/17
B = 3/17

So, Y(t) = (-5/17)sint(t) + (3/17)cos(t)

Lets try one more,

y'' - 3y' - 4y = 4t2 - 1

Since g(t) is a polynomial with terms t2, t, 1, with coefficients 4, 0, -1 respectively, we can assume the solution;

Y(t) = At2 + Bt + C
Y'(t) = 2At + B
Y''(t) = 2A

[2A] - 3[2At + B] - 4[At2 + Bt + C] = 4t2 - 1

[-4A]t2 + [-6A - 4B]t + [2A - 3B - 4C] = 4t2 - 1

-4At2 = 4t2
A = -1

-6At - 4Bt = 0t
-6(-1) = 4B
B = 3/2

2A - 3B - 4C = -1
2(-1) - 3(3/2) - 4C = -1
-4C = 11/2
C = -11/8

So, Y(t) = (-1)t2 + (3/2)t + (-11/8)

Problems:

Find the general solution of;
1. y'' - 2y' - 3y = 3e2t
2. 2y'' + 3y' + y = t2 + 3sin(t)

3. y'' + 2y' + 5y = 4e-tcos(2t), y(0) = 1, y'(0) = 0

[hawaiidude]

yeah thanks....by the way, how do you compute pde's? the advanced types?

[ExtravagantDreams]

I believe that is Fourier series, something I will not learn until Monday.

[hawaiidude]

i thought pde's were partil differential equations>?

[selfAdjoint]

He meant that's one way to solve them, I think. Yes PDE means partial differential equation. Some of them can be reduced to ordinary differentials and there are just tons of methods for particular cases. This has been one of the most active branches of math resarch for hundreds of years, and no end in sight.

[hawaiidude]

o..by the way..here are some things that i wish to know
wha are the recurrence formulass for the folowwing

1) (x^2 +4) y''+xy=x+2
2) y''+y=0
8x^2 y''+10xy'+(X-1)y=0

[ExtravagantDreams]

We have already seen how to find a particular solution for nonhomogeneous equation using the Method of Undetermined Coefficients, now we will try to use variation of parameters to accomplish the same thing.

Let's jump straight to an example;

y'' + 4y = 3csc(t)

Noting that the corresponding homogeneous equation is;

y'' + 4y = 0

We first solve this equation.

r12 = &radic;(-4(1)(4))/2(1) = 2i

remember that solution will be in form;

e&lambda;t[cos(&mu;t) + sin(&mu;t)]

where in this case &lambda; = 0 and &mu; = 2, so

yc(t) = c1cos(2t) + c2sin(2t)

The basic idea is to replace the constance c1 and c2 with functions u1(t) and u2(t) and solve for these functions.

Starting with the equation;

y = u1(t)cos(2t) + u2(t)sin(2t)

we can differentiate to optain;

y' = u'1(t)cos(2t) + u'2(t)sin(2t) - 2u1(t)sin(2t) + 2u2(t)cos(2t)

Since we only have one initial condition so far, yet two unknown variables, this would give us infinite many solutions. Let us impose a second condition so that we have one final solution. Here it is not important why we can do this;

We require that;

u'1(t)cos(2t) + u'2(t)sin(2t) = 0, so;

y' = 2u2(t)cos(2t) - 2u1(t)sin(2t)

y'' = 2u'2(t)cos(2t) - 2u'1(t)sin(2t) - 4u2(t)sin(2t) - 4u1(t)cos(2t)

Substitude these equations back into the original equation;

[2u'2(t)cos(2t) - 2u'1(t)sin(2t) - 4u2(t)sin(2t) - 4u1(t)cos(2t)] + 4[u1(t)cos(2t) + u2(t)sin(2t)] = 3csc(t)

2u'2(t)cos(2t) - 2u'1(t)sin(2t) = 3csc(t)


From our second set condition;
u'1(t)cos(2t) + u'2(t)sin(2t) = 0

u'2(t) = -u'1(t)cos(2t)/sin(2t)


Substitude;
2[-u'1(t)cos(2t)/sin(2t)]cos(2t) - 2u'1(t)sin(2t) = 3csc(t)

Simplify;

u'1(t) = -(3csc(t)sin(2t))/2 = -3cos(t)

Substituging once more;

u'2(t) = -u'1(t)cos(2t)/sin(2t)
u'2(t) = [-3cos(t)]cos(2t)/sin(2t)
u'2(t) = (3/2)csc(t) -3sin(t)

Now that we have optained u'1(t) and u'2(t), Integrate;

u1(t) = -3sin(T) + c1
u2(t) = (3/2)ln|csc(t) - cot(t)| + 3cot(t) + c2(t)

Finally, substitude u1(t) and u2(t) into the y expression;

y = [-3sin(T) + c1]cos(2t) + [(3/2)ln|csc(t) - cot(t)| + 3cot(t) + c2(t)]sin(2t)


That probably looked more confusing than need be, so lets look at an arbitrary function to see a set by set method and prove this can be used for any Second Order Linear Nonhomogeneous Equation.

Let us start with the general equation;
y'' + p(t)y' + q(t)y = g(t)

The general solution to the corresponding homogeneous equation will be;
yc(t) = c1y1(t) + c2y2(t)

This is from the assumption that the equation has constant coefficients. Now in the general solution, replace constants with functions u.

y = u1(t)y1(t) + u2(t)y2(t)

Take the derivative;
y' = u'1(t)y1(t) + u'2(t)y2(t) + u1(t)y'1(t) + u2(t)y'2(t)

For a second condition set terms with u' equal to zero;
u'1(t)y1(t) + u'2(t)y2(t) = 0

This gives;
y' = u1(t)y'1(t) + u2(t)y'2(t)

Differentiate again and plug y, y', and y'' into the original equation;

y'' = u1(t)y''1(t) + u2(t)y''2(t) + u'1(t)y'1(t) + u'2(t)y'2(t)

[u1(t)y''1(t) + u2(t)y''2(t) + u'1(t)y'1(t) + u'2(t)y'2(t)] + p(t)[u1(t)y'1(t) + u2(t)y'2(t)] + q(t)[u1(t)y1(t) + u2(t)y2(t)] = g(t)

Rearranging;
u1(t)[y''1(t) + p(t)y'1(t) + q(t)y1(t)] + u2(t)[y''2(t) + p(t)y'2(t) + q(t)y2(t)] + u'1(t)y'1(t) + u'2(t)y'2(t)] = g(t)

Since both y1 and y2 are solutions to the corresponding homogeneous equation, the expressions in brackets equal zero, leaving;

u'1(t)y'1(t) + u'2(t)y'2(t) = g(t)

Using this equation and the previous equation;
u'1(t)y'1(t) + u'2(t)y'2(t) = 0

substituation can be used and integration can be done to find u1and u2.

u'1(t) = -y2(t)g(t)/W(y1,y2)(t)

u'2(t) = y1(t)g(t)/W(y1,y2)(t)

u1(t) = -&int;(y2(t)g(t)/W(y1,y2)(t))dt + c1

u2(t) = &int;(y1(t)g(t)/W(y1,y2)(t))dt + c2

Where then;
Y(t) =
-y1(t)&int;(y2(t)g(t)/W(y1,y2)(t))dt + y2&int;(y1(t)g(t)/W(y1,y2)(t))dt

and the general solution is;
y = c1y1(t) + c2y2(t) + Y(t)

I realize this is a little confusing to follow, so let me sum up what you really need to know without deriving everything everytime.

First;
You must find the solutions y1 and y2 of the homogeneous equation.

Then use these two formulas:

u'1y1 + u'2y2 = 0
u'1y'1 + u'2y'2 = g(t)

If there is a term infront of the y'' term, it must be divided out to give the correct g(t).

From this system of equations, where y and y' are known, u'1 and u'2 can be found, then integrated.

[QuantumNet]

DDy + Dy + y = 0 for all values of y

x2 + px + q cannot be zero for all values.

Although it's possible if y = aekx.

y = aekx
Dy = akekx.
DDy = ak2ekx.

DDy + Dy + y = ak2ekx + akekx + aekx = a(k2 + k + 1)ekx

If (k2 + k + 1) = 0 then DDy + Dy + y = 0

[ExtravagantDreams]

Awsome, I feel special now that this was made a sticky [6)]

I haven't been around in a while, I was really busy during finals and then kinda crawled in a hole for a month during break. but for now I am back, we will see how long it lasts this time.

I changed my format of doing this, instead of writing everything out on here, I have decided to use Word Processor and Equation Editor to create a document that you can then download. I hope no one will have any problems this way.

If you dare... (http://www.angelfire.com/tx5/extravagantdreams/Series_solutions_of_Second_Order_Linear_Equation.d oc)

[nickdanger]

However, a question, does anyone know an easier way for writing math on the computer and one that looks less confusing. I know I will have difficulty finding some things, especially subscripts and superscripts. Anyone know a better way to denote these?

My understanding is that PF postings will support latex formatting. I haven't used it yet but go to: http://physicsforums.com/showthread.php?t=8997
for instructions.

[Ebolamonk3y]

Does anyone know of any good Intro to Diffy Q books? Or just Diffy Q books in general? Thanks...

[abacus]

Does anyone know of any good Intro to Diffy Q books? Or just Diffy Q books in general? Thanks...


I'm using the book Differential Equations: 2nd Edition by Blanchard, Devaney, Hall.

[Dr Transport]

Try Boyce and DiPrima.......it hasn't been thru 7 or 8 editions beause it is not a good, readable text

[Tom Mattson]

Try Boyce and DiPrima.......it hasn't been thru 7 or 8 editions beause it is not a good, readable text

I second that. I actually took the Diff Eq with Boyce--great teacher.

The Schaum's outline "Modern Introductory Differential Equations" is also very good.

[tavi_boada]

Apostol's calculus is great

[zeronem]

Im going to begin by implicit differentiation, then I will ask the question I have. The question is on finding the interval where solution is valid.

xy^2 - e^{-y} - 1 = 0

u = x, v = y

(u + du)(v+dv)^2 = (u + du)(v^2 + 2vdv + d^2v)

= uv^2 + 2uvdv + ud^2v + v^2du + 2vdvdu + d^2vdu

we get rid of uv^2 and any part of the equation that contains more then one differential. Lets say w = uv^2, then w + dw = the above equation. dw would equal the above equation without the uv^2 since you subtract the w. You get rid of the parts of the equation that has more then one differential because more then one differential is just simply too small to have any effect on the whole equation. We then get

2uvdv + v^2du

We then plug in for x and y and get

2xyy' + y^2

We then have implicitly differentiated part of the equation. We now continue to implicitly differentiate the rest of the equation.

z = -y

-e^{-y} = -e^z

dz = -dy

\frac {d(-e^z)} {dz} = -e^z = -e^{-y}

we then take dz and multiply it by \frac{d(-e^z)}{dz} ,

(dz)(-e^{-y}) = + e^{-y}y'

we then differentiate -1 which becomes a 0 and we get

2xyy' + e^{-y}y' + y^2 = 0

As implicit differentiation of xy^2 - e^{-y} - 1 = 0

We then test the implicit differentiation of the above equation as the solution to the differential equation

(xy^2 + 2xy - 1)y' + y^2 = 0

We do this by solving for e^{-y} from the equation xy^2 - e^{-y} - 1 = 0

e^{-y} = xy^2 - 1

we plug e^{-y} in the implicit differential equation and get (2xy + xy^2 -1)y' + y^2 = 0

So it is a solution of the differential equation (xy^2 + 2xy - 1)y' +y^2 = 0

Now this is where I need help. Can I say that the original equation xy^2 - e^{-y} - 1 = 0 is an implicit solution to (xy^2 + 2xy -1)y' + y^2 , even though it works out?


Exactly how do I find such interval where the y is an implicit function of x on an interval. So I can determine whether the equation xy^2 - e^{-y} - 1 = 0 is an implicit solution to (xy^2 + 2xy -1)y' + y^2

[Fritz]

\frac {dy} {dt} = ay - b



Here is y the independent variable and a and by the dependent variables?

[QuantumTheory]

You get rid of the parts of the equation that has more then one differential because more then one differential is just simply too small to have any effect on the whole equation.1 = 0 [/tex] is an implicit solution to ..

YES, YES, YES!!

This is a problem I am having!

I understand that alot of people here dislike the book Calculus For Dummies, for beginners. Alot of people in the real world cannot understand calculus, it is hard. This book really opened my eyes. I understood that a derivative if nothing but a constant rate. I learned the basics to move on.

However, before I got this book, I picked up my first calculus book in my school library. It was made in the very early 1900's, and was about calculus. Beginner calculus.

In this book, he actually had equations of multi-varibles!

For example, dx^2/dy^2, or anything with an infinitely small piece of a deriviative squared.

Since a deriviative is an infinitely small section, out of a whole of an infinite amount of infinitely small pieces, how could you have basically a negative infinity squared? This just doesn't make sense. You're already at some negative infinity rate of one dx. So if the derivative is the rate of some curve(therefore the height of the curve), lets call it f(t).

Then,

f(t) = f(t)dx


Does this make sense? I made it up myself..I hope its right.

Basically, I don't understand how you could possibly have some derivative, of some function, squared.

Ugh, anyway, I hate infinity, it confuses me sometimes :(

Weird how an infinite amount of infinitely thin pieces can equal a finite amount, eh?

[poolwin2001]

The link is not working.

[whozum]

I have to say this guide is amazing. I have one question on post 25:

py'' + qy' + ry = 0

Solving the homogeneous equation will later always provide a way to solve the corresponding nonhomogeneous problem.

I'm not going to proove all this but you can take the kernal of this funtion as

ar2 + br + c = 0

and you can, so to speak, find the roots of this funtion.

r1,2 = (-b ± √(b2 -4ac))/2a

r1 = (-b + √(b2 -4ac))/2a
r2 = (-b - √(b2 -4ac))/2a

Assuming that these roots are real and different then;
y1(t) = er1t
y2(t) = er2t


How did you get from the above to the last statements?

[dextercioby]

Which one,the first statement?

Daniel.

[whozum]

How does:


Assuming that these roots are real and different then;
y1(t) = er1t
y2(t) = er2t


follow from


py'' + qy' + ry = 0

Solving the homogeneous equation will later always provide a way to solve the corresponding nonhomogeneous problem.

I'm not going to proove all this but you can take the kernal of this funtion as

ar2 + br + c = 0

and you can, so to speak, find the roots of this funtion.

r1,2 = (-b ± √(b2 -4ac))/2a

r1 = (-b + √(b2 -4ac))/2a
r2 = (-b - √(b2 -4ac))/2a


I think i understand the root analogy and how r1 and r2 come about, but I dont understand why y1 and y2 follow that structure.

[dextercioby]

That's how u got the characteristic equations,by assuming exponential type solutions and plugging in the ODE.

Daniel.

[Maxos]

Yes, you make the hypothesis:

Y(t)=e^(a*t)

then, if Y(t) must follow the diff.eq. for any real t

e^(a*t)*(p*a^2+q*a+r)=0

from this You get the two (or one) values.

Then, for some algebrical reasons, you demonstrate that the general integral is a linear combination of the found solutions.

[jaap de vries]

Does anyone know an easier way for writing math on the computer and one that looks less confusing. I know I will have difficulty finding some things, especially subscripts and superscripts. Anyone know a better way to denote these?

Just use _ and ^ in LAtex its pretty easy

[Ruslan_Sharipov]

Looking things up and explaining it to others seems to be the best way to learn.


I would like to suggest you something more than to learn a standard textbook on Diff. Equations. Great many of the ODE's you can find in applications are special cases of the following one:

y''=P(x,y)+3 Q(x,y) y'+3 R(x,y) (y')^2+S(x,y) (y')^3,

Once you have such an equation, you usually try to reduce it to some standard equation given in a reference book. A change of variables is most often used for this purpose:

\tilde x=\tilde x(x,y),


\tilde y=\tilde y(x,y).

Such a change of variables is called a point transformation. There is a theory giving some hints how to find proper point transformation for a given equation. It would be best if you learn this theory and explain it to others (including your class instructor). The following links will help you:
http://arxiv.org/abs/math.DG/9802027
http://arxiv.org/abs/solv-int/9706003

Best regards,
Ruslan Sharipov (mailto:r-sharipov@mail.ru)

[senthanal]

Please send this docment.

[nazgjunk]

Hello all, I am pretty new to these forums, so correct me if I do anything wrong.

In maths class, we have just finished a chapter on differential equations. However, we were mainly working on something like this:

f(x)=2x^4+x^3-0.2x^2+3x
*snip*
f'(x)=8x^3+3x^2-0.4x+3

Now I am wondering why I understand so little of what this thread is saying. I figure there are some possibilities:
1: The way of writing down these things is way different from what I have learned

2: I haven't learnt very much yet. At school we have come as far as the multiplication and division rules, eg:
f(x)= \frac {g}{h}

f'(x)=\frac {g'h-h'g}{h^2}


3: I am completely wrong in translating.

Any hints, please? Because I do like to learn.

Thanks in advance,

Nazgjunk

[hypermorphism]

Hello naz,
The equations you wrote down are derivatives. In other words, given a function, those formulas tell you how to find a derivative.
The study of differential equations is the study of how one can get a set of functions that satisfy a given derivative (It will always be more than one because derivatives annihilate additive constants).
For example, if you know f(x) = - f'(x), we want to find some explicit form of f(x) that will give us that equation as a derivative. In this case, a moment's thought will show f(x) can be of the form f(x) = A*cos(x + B), where A and B are arbitrary constants.
Usually in physics one first sees the differential form of some phenomenon (ie., simple harmonic motion) and one then tries to solve the differential equation(s) to get a more explicit form.

[nazgjunk]

Ok, thanks. I think I got the basic idea, and it indeed is probably a translation problem. I still think it's weird, though: my dictionary says the Dutch "differentieren" does mean "differentiate", but it seems to be something completely different.

[mathwonk]

boy am mi ticked. i just lost a post that I had been woprking nop fopr over an hour about o.d.e's from the biog picture and various books and their different characteristics, and essential ingredients of a good d.e. cousre etc etc. when i tried to pst it the computer said I was not logged in but when I logged in my post was gone.

this is not the first time this has happened to me.

well good luck for you, bad luck for me.

[shmoe]

boy am mi ticked. i just lost a post that I had been woprking nop fopr over an hour about o.d.e's from the biog picture and various books and their different characteristics, and essential ingredients of a good d.e. cousre etc etc. when i tried to pst it the computer said I was not logged in but when I logged in my post was gone.

this is not the first time this has happened to me.

well good luck for you, bad luck for me.

I have to say this is a shame. Your "big idea" posts where you run down the main ideas of a topic are some of my favorites.

If I understand what you did, you typed the message, then hit submit but had to log in and then it gave you a blank form (or invalid thread)? This happens to me all the time, hitting the "back" button a couple times usually gets me back to the message I had typed, and it's recovered. Paranoia also makes bme sometimes "copy" a long message before I hit "submit".

[Maxwell]

boy am mi ticked. i just lost a post that I had been woprking nop fopr over an hour about o.d.e's from the biog picture and various books and their different characteristics, and essential ingredients of a good d.e. cousre etc etc. when i tried to pst it the computer said I was not logged in but when I logged in my post was gone.
this is not the first time this has happened to me.
well good luck for you, bad luck for me.

When you type up a big post, copy* it before you submit it. I've been burned too many times before to ever let a huge post of mine disappear because of some bad message board voodoo!

*Select your whole post and press CTRL V, incase you didn't know.

[mathwonk]

It happened again, but I SAVED!! Here is todays post:
I am starting to teach o.d.e. and the first thing I am going to do tomorrow (the 2nd day of class) is try to explain why the sort of thing appearing in post #3 here, apparently taken from Boyce and Diprima, (and not to be blamed on the student trying to learn it, my apologies to that student) is completely meaningless nonsense.
I.e. solving the d.e. "dy/dt = g(t)" may or may not be possible, depending on the nature of g.
In particular it makes no sense at all to simply write
"indefinite integral of g = y", and claim to have solved the problem, since the notation "indefinite integral of g" stands for any function whose derivative is g, provided one exists.
So one has made no progress at all in solving the problem in writing this, as one is merely restating it in different notation. Books which say this sort of thing drive me nuts, as they give young students entirely the wrong idea as to what a function is, and what it means to "solve" a d.e.
Assuming "solving" an equation means "finding" a function that satisfies it, what does it mean to "find" such a function? To call out its name? If so, then I have solved dy/dx = 1/x by saying the magic words "natural log". Unfortunately I do not know the value of this function even at the number t = 3.
So I really do not know much about this solution except that it exists.
Actually I don't even know that, I am only claiming it exists, since most students probably cannot really prove the natural log function does exist, and satisfies this equation. But I have been told it does, so I say "natural log" solves this equation.
But wouldn't it make more sense to say I have "found" a function, if I can actually tell you some of its values, or even an arbitrary value, at least to any desired degree of accuracy?
The way to do this with ln(t) is actually to approximate the area under the curve of y = 1/t. Defining ln as that area fucntion at least shows it exists, but I still ought to prove that area function is both differentiable and solves the equation.
So for some g, an antiderivative function exists and for others it does not. For example it does exist for g(t) = 1/t, or g(t) = e^(t^2)), but does not exist for g(t) = 0 for t<0 and g(t) = 1, for t g.e. 0.
The usual sufficient criterion for such an antiderivative to exist is that g be continuous. But these hypotheses are nowhere mentioned, before writing "indefinite integral of g".
Indeed the indefinite integral of 1/t does not exist because ln(t) exists, rather it is the other way around, ln(t) exists for all t>0, because the area function of 1/t exists, because 1/t is continuous.
So the area function (definite integral) is a machine for making differentiable functions out of continuous ones. Every now and then we will learn that some area function equals some other function we have met in a different situation, but so what?
The ones we have met before are no better solutions than the ones we have not. We just understand those better. So books that say: "well, we can solve the d.e. dt/dy = 1/t, but not dy/dt = e^(t^2), because e^(t^2)) cannot be integrated", are lying, and promulgating a false understanding of d.e's, functions, and their solutions.
More correct would be to say: every continuous integrand leads us to a differentiable area function, and to a potentially new and interesting function. A few of these we have met before and given names to.
But in general the integral (area function of) a function is more complicated than the original fucntion, and many of them we have not yet had time to name.
E.g. the integrand g(t) = 1, has area function t+c, and the integrand g(t) = t^r has area function (1/r+1) t^(r+1), except when this makes no sense, namely when r = -1.
In that case the area function (starting at t=1) of 1/t is a function which we shall call ln(t). It happens to be inverse to an exponential function e^t with a base "e" which we would never have met if we were not solving this d.e. But otherwise it resembles exponential functions like 2^t which of course we previously only defined for rational t.
Now this lets us also name the area functions for all fractions, since by partial fractions (using complex numbers) all fractions have integrals which reduce to sums of ones like 1/(t-a), and these are also all natural logs. So spending hours and weeks integrating more and mroe examples of fractions is just repeating the same thing over and over, or if you like, it is practicing algebra. But once you have integrated 1/(t-c) you are not learning anymore about integration by integrating 1/(any polynomial in t).
Now lets try some really new integrands (studied by the Bernoullis in the 1600's?). lets put a square root in the denominator.
e.g. let g(t) = 1/sqrt(1-t^2). this is continuous within the interval -1<t<1, so has a differentiable area function there, which turns out to be the arclength function for a circle, and it already has a name "arcsin" because arclength for circles came up long ago. (Euclid, 1800 BC?)
Moving on we try perhaps 1/sqrt(1-t^3), also continuous on (-1,1) so it too has a nice differentiable area function but most of us do not know any name for it. But Weierstrass studied it in the 1900's and it is called an elliptic integral because it comes up in trying to measure arclength on a lemniscate? Wait a minute, for this story to be any good, it should be arclength on an ellipse. Well maybe it comes up there too.
Now shall we say this integral does not exist? or that we have not solved the d.e. dy/dt = sqrt(1-y^3) ? just because we have not yet chosen a name for this function?
That would be absurd. so we call it maybe frank, as one of my friends says.
anyway, it should be obvious that all the integrands 1/sqrt(1-t^n) do have differentiable area functions, and all deserve names, but we only name the ones we need in our own problems.
frank turns out to have an inverse function which is periodic like sin and cos, but even better, it is not just singly periodic, but doubly periodic in the complex plane. These functions are really wonderful, and not only do exist, but played the main role in Wiles solution of Fermat's last theorem.
So one should never give students the idea that those d.e.'s whose solutions are functions we have not named yet, or whose names we have not heard yet, somehow are not solved, or even not solvable, and yet this is precisely the impression i get from some books.
Solving an equation means producing a function that solves it. Producing a function means defining it, not just hollering out its name (Oh yeah, I know the solution of that equation, it's Harold, or is it Maude? Is that any worse than a student saying the solution of dy/dt = 1/(1+t^2) is arctan, but the student does not know one single value of arctan?).

Defining it can be done by a lot of different processes, usual among these being "taking the area function of a given function" i.e. really integrating it, or inverting a given function.
Now here is the first lesson of o.d.e.:
If g(t) is any continuous function on an interval I, then the area function of g (Riemann's definite integral of g from a to t) is a differentiable function on I which solves the d.e. dy/dt = g(t).
second lesson of o.d.e.:
If g(t) is any continuous function that is never zero on an interval I, then the area function of
1/g(t) is an invertible differentiable function on I, whose inverse function solves the o.d.e. dy/dt = g(y). (That's right the letter in g is y not t, and the g moved from the denominator to the numerator, because that's what happens to the derivative when you take an inverse function.)
This method is usually called "separation of variables", or just "integration".
I.e. the "monkey see monkey do" way of solving the d.e. dy/dt = g(y)
is to multiply by dt and divide by g(y) and get dy/g(y) = dt, thus "separating the variables".
Then, step one, "integrate" both sides, to get G(y) = t+c. (*)
then step 2: invert the function G (which has an inverse because its derivative 1/g(t) was assumed to be zero nowhere), getting a function H,
and
step 3): apply H to both sides of (*) getting: y = H(t+c).
Now none of this makes any sense, unless you understand how the two processes, taking area function, and inverting a function, do in fact transform an appropriate differentiable function, i.e. one satisfying certain hypotheses, into another differentiable function, namely the solution.
ok, I iterate: the student learning this material, or trying to, from the usual books is not to blame for the confusion the books spew everywhere, but is actually the victim.
My point is: one has NOT solved an o.d.e. simple by saying "the name of the solution is frank", but by knowing why the solution exists, what its domain is, what properties it has there, and how to approximate its values there, and sketch the graph.
E.g. one can do all of these things for the solution of dy/dt = e^(t^2)), so it is quite false for books like the one I was reading today, to say "one cannot integrate this d.e. directly".
ok, end of rant. I am just finding out why I never understood this subject, as the books on introductory d.e. are probably the worst in all of elementary mathematics. it is very hard to find a decent, correct, explanation of d.e. the best I have found is the book of v.i. arnol'd, and that is pretty dense going. I admit also to having learned something from Blanchard Devaney and hall, but it is very wordy and there is no theory at all.
Most of the rest are cookbooks of the worst sort, teaching you to spout out the usual names of the solutions to the simplest possible equations,
i.e. ones like y''' + 3y'' + 3y' + y = 0, which are just chosen because they are all (complex) exponentials (i.e. cosines and sines and exponentials).
A d.e. is a racecourse, with speed signs all over, and a solution is a driver in a fast, well handling car, navigating the course at the right speed at every instant, and in the right direction.

[mathwonk]

bvy the way, there is essentially nothing to solving all d.e's of form:

dy/dt = P(y) where P is any polynomial, since these are all solved by first integrating 1/P(t), and then inverting the answer. (i.e. separating variables).

Similarly, all equations of form ay^(n) + by(n-1) + cy^(n-2) + .....+ey = 0, are all compositions of the equation dy/dt = ay, so are all solved by the same idea as for solving that one. In fact they can be considered as that same equation in the form dY/dt = AY, where Y is a vector and A is a matrix, and then the solution is actually just the vector Y = e^(At), where e^(At) is defined by the same power series as e^(at).

I guess I could call this the 3rd lesson of o.d.e.

In spite of the fact these are almost all the same equation, namely the world's easiest one: dy/dt = a(y-c),

one can spend a whole semester on them, and many books spend hundreds of pages on them.

[mathwonk]

OK I just read the whole thread and it looks really nice, my compliments to extravagant dreams. you clearly know more about d.e's than me.

still i am bold enough to offer my views.

I liked post #20, where the existence theorem is discussed. this is really important, since it shows how to solev ALL such d.e.'s. namjely you find a way of approximating a solution, then you find a process of amking the approximations betterm then you prove that in the limit they converge to a perfect solution.


This is actually useful when needing only an approaximate solution, i.e. in real life.


then post #39 interested me since mit shows the limitations of this books approach, namjely not explaining what is going on and just telling the student to try thios and that.


the point is that: 1) the non homogeneous equation can always be solved by varying the solution of the homogeneous equation.

then they are limiting their study to thoe homogeneous equations which are basically just variations on dy/dt = y, so the solutions are always exponentials and sins and cosines, which we know are the same as (complex) exponentials.


But then theory of even these equation is missing:

look at this equation: y'' - 3y' - 4y = sin(t).


The point is to look at the elft hand side as a linear operator on functions, just like in linear algebra, i.e. L(y) = y'' - 3y' - 4y. Now we want ay such that L(y) = sin(x).


First of all, we should solve the equation L(y) = 0, to find the kernel of the l;ienar operator.

now to do thsi correctly, instead of just memorizing the characteristic equtions approach, one should factor the linear operator L as

L = MoN where M,N are simpler differential operators.

I.e. let D be the simplest differential operator of all, D(y) = y'.

then let M = D+1, so M(y) = (D+1)y = y'+y.

And let N = (D-4) so N(y) = (D-4)y = y' -4y.

Then L(y) = (D+1)(D-4)y = (D^2 -3D -4)y = y'' - 3y' - 4y.

Thus if we compoute the kernel of M, then the kernel of L = NoM, is the inverse imnage of the kernel of M under the operator N.

I.e. we know that M(y) = (D+1)y = y'-y = 0 iff y' = -y, iff y = ce^(-t).


so the kernel of M is {ce^-t} for all c.

then we need to know which y's are such that Ny = ce^-t?


So solving the homogeneous equation My = 0, is the same as solving the non homogeneous equation Ny = ce^t.


But, N kills de^4t, and it is not too hard to guess that N acts on 3^-t to give something of form e^-t. of course this is also guessing but it is easier guessing.

i.e. we get, N(e^-t) = (D-4)e^-t = -e^-t -4e^-t = -5e^-t.

spo that can easily be solved for something that maps straight to e^-t, namely N(-1/5 3^-t) works I guess.


So now since the kernel of N is de^4t, and N maps (-1/5 3^-t) to e^-t, I guess, oh yes it is even easier than I made it,

N maps things of form ae^4t + b e^-t to things of form ce^-t.

Thus L = NoM kills things of form ae^4t + b e^-t.


(maybe I got the order backwards, but it does not matter as NoM = MoN, i.e. these operators commute.)


Now this decomposition idea also makes the non homogeneous problem easier and more motivated.


I.e. now lets attack y'' - 3y- -4y = sin(t).

well first we should look at the composition NoM(y) = N(M(y)) = sin(t), so we need something that N takes to sin(t), so we need to solve

N(y) = sin(t). Now it becomes really clear why sin(t) does not work, because

N(sin(t)) = -4sin(t) + cos(t). so cloearly you are going to need some sins and cosines to do this. this seems a more natural way to see why the solution should like like asin(t) + bcos(t).

of course then once we have an f such that (D-4)f = sin(t), we still need to find a g such that (D+1)g = f. This si the same process.

Now it looks as if the solcvinbg process is idneed shorter when done as in the post 39. But thinking of it seems more naturla to me anyway, this way.

I.e. bring in some linear algebra.

[mathwonk]

By the way, Sharipov's post is to me a nice remark that this book being studied is just going over and over the same d.e., namely variations on dy/dt = y, so one is not getting too far, and it might be useful to branch out a bit.
i.e. what has been done here is to remark that linear equatioins can be solved by variation of parameters, provided the homogeneous one can be solved. then atgtention is restrictred tot he constant coefficient case, where the homogeneous one is always solved by exponentials.
the one extension of that theory is a nice excursion into power series, by showing that with non constant coefficients, one can at least solve recursively for the power series solution of other linear equations.
the other idea is that in dealing with linear equations one can always solve for the sum of any functions one can already solve for. This makes sense also for infinite sums. hence one can solve for infinite sums of sines and cosines i.e. exponentials. these are called fourier series methods.
so basically all linear (constant coeff) d.e.s are solved by exponentials.
the higher order ones are no different if one uses matrix exponentials.
I.e. to solve y'' - 3y' -4y = 0, look at it as a 2x2 system, for a vector function of t, namely (x(t), y(t)) where x = y' and x' = y'' = 3y' +4y = 3x+4y.
so the linear system is (x,y) where (x',y') = (3x+4y,x).
Thus our matrix is A = with first row [3,4] and second row [1,0]. the characteristic polynomial of A is (surprize surprize) x^2 - 3x - 4, with eigenvalues x = -1, 4.
So we can diagonalize the matrix as A' = diagonal matrix with -1,4 on main diagonal.
then the exponential of the matrix tA' is just the diagonal matrix with
e^-t and e^4t on the diagonal. the eigenvectors are (1,-1) and (4,1), and the solution vectors of the system are linear combinations of (ce^-t, -ce^-t), and (4d(e^4t), de^(4t)).
In particular the solutions of the original equation, i.e. the y entry of such a linear combination, are of form ae^-t + be^4t.
but what if we want to understand a pendulum, i.e. find y satisfying the equation:
d^2y/dt^2 - sin(y) = 0?
we need some more tools. so we should teach phase plane analysis, i.e. how to draw pictures of vector fields and pictures of the solution flows.

[mathwonk]

another topic that is often omitted, but that arnol'd clarifies, is that of the domain of the solution. i.e. if the manifold where then vector field is defiend is compact, then the domain of the solution is all of R, so one gets a 1 parameter flow on the whole manifold, i.e. any point can be flowed along a solution curve for time t, whatever t is. this adds much to the geometry of the subject.

[total chaos]

guys,was just reading diff.eqns by ross.
uniqueness of an 1-d d.eqn of the form
y'=f(x,y) is said to be guaranteed when
i>partial der.of f(x,y) w.r.t y is a continuous of x & y over the same domain D over which f(x,y) is defined & is continuous.
can somebody explain how this condition guarantees uniqueness of the soln??

[HallsofIvy]

guys,was just reading diff.eqns by ross.
uniqueness of an 1-d d.eqn of the form
y'=f(x,y) is said to be guaranteed when
i>partial der.of f(x,y) w.r.t y is a continuous of x & y over the same domain D over which f(x,y) is defined & is continuous.
can somebody explain how this condition guarantees uniqueness of the soln??

In order that the differential equation, dy/dx= f(x,y) have a unique solution (in some region around x0) satisfying y(x0)= y0 it is sufficient but not necessary that f be differentiable with respect to y in some region around (x0,y0. It is sufficient that f(x,y) be "Lipschitz" in y: If (x, y1) and (x,y2) are two points in that region, then |f(x,y[sub]1[sub])- f(x,y2)|< C|y1- y2|. If a function is differentiable in a region, then you can use the mean value theorem to show that it is Lipschitz in that region but there exist functions that are Lipschitz in a region but not differentiable there. Most elementary texts use the simpler "differentiable" as a necessary condition and may or may not point out that it is not sufficient.

Picard's proof of the existence and uniqueness of such a solution is long and deep. I won't post it here, but this is a link to a detailed explanation:

http://academic.gallaudet.edu/courses/MAT/MAT000Ivew.nsf/ID/918f9bc4dda7eb1c8525688700561c74/$file/Picard.PDF

Essentially, you replace the differential equation by the corresponding integral equation (dy/dx= f(x,y) if and only if y= \int f(x, y(x)) dx) then show that the integral operator defined in that equation satisfies the hypotheses of Banach's fixed point theorem.

[total chaos]

sorry for responding after such along time. 'was sick & hence away from my institute.
thanx halllsofivy, but i did not ask how to prove that continuity of p.der of f wrt y was a sufficient condition, rather i wanted to know if it is possible to EXPLAIN this requirement in simpler terms !!
let me explain,existence of a soln of the problem of the type given by me above is guaranteed if there f(x,y) is a cont fn of x& y in some domain D.this can be explained thus:: if f(x,y) was not cont, it wud not have been possible to integrate f(x,y) over the domain D (which is after all the method by which we find out the soln).
what i'm seeking is some explanation of this nature!!
thanx in advance!!

[HallsofIvy]

but i did not ask how to prove that continuity of p.der of f wrt y was a sufficient condition, rather i wanted to know if it is possible to EXPLAIN this requirement in simpler terms !!
No!!!

let me explain,existence of a soln of the problem of the type given by me above is guaranteed if there f(x,y) is a cont fn of x& y in some domain D.this can be explained thus:: if f(x,y) was not cont, it wud not have been possible to integrate f(x,y) over the domain D (which is after all the method by which we find out the soln).
??Why not? The fact that a function is not continuous does not mean it is not integrable. Even if that were true that would be showing that continuity is a necessary condition, not a sufficient condition, which is what by "solution is guaranteed". If that's the kind of "explanation" you are looking for, I surely can't help you!

[bioginz]

Hey Guys,

I have learned that Mars and Earth closest encounter so far happened August 27... and it is estimated to commence again by year 2287 if im not mistaken.... I am just wondering how the calculations were made employing the principles of Differential Equations? Im really curious 'bout the accuracy or exactness of the date or just the exact year only...

Hoping for your prompt response.

DANDYBOY

[sina_box]

hi to everyone
I am a new user and happy to find this forum.so i'm looking for a way to solve these equations by using delphi programming language.i am a physics student and have "use computer in physics" in this course.
thank you

[mrmotobiker]

i just found this useful topic to use in my DE class. does anyone know where i could find solutions (the ones located on the first page of this topic)? they are all too old or not available.

[LRJ85]

hey thanks...nice examples,,,clear and easy to understand...but heres a problem...when x=0 3x^2y''-xy'+y=0

I tink what hawaii meant was to find the solution to the diff. eqn 3x^2 \frac{d^2y}{dx^2} - x \frac{dy}{dx} + y = 0 at the point xo = 0. After looking through the "Series solution of 2nd Order Linear Equations" in thread #47 written by ExtravagantDreams, i realize that the above diff. eqn can be solved easily. The solution will be of the form y = \sum_{n=0}^\infty a_n(x - x_0)^n .
Since xo = 0, y = \sum_{n=0}^\infty {a_n x^n} .
y' = \sum_{n=0}^\infty {a_n nx^{n-1}} .
y'' = \sum_{n=0}^\infty {a_n n(n-1)x^{n-2}} .
Now, substitute these expressions into the original diff. eqn:
3x^2 \sum_{n=0}^\infty {a_n n(n-1)x^{n-2}} - x \sum_{n=0}^\infty {a_n nx^{n-1}} + \sum_{n=0}^\infty {a_n x^n} = 0
Factor in the external x terms:
\sum_{n=0}^\infty {3a_n n(n-1)x^n} - \sum_{n=0}^\infty {a_n nx^n} + \sum_{n=0}^\infty {a_n x^n} = 0
Combining all the terms since they already have the same degree and same starting pt:
\sum_{n=0}^\infty {[3n(n-1) - n + 1]a_n x^n} = 0
So finally we arrive at 3n² - 4n + 1 = 0 and an = 0. We get 2 values of n, n=1 and 1/3. How do i proceed from here then??

Any good help would be appreciated :)

[LRJ85]

ok heres question.... solve the differential equation d^3x/dt^3-2d^2d^2x/dt^2 +dx/dt=0 and find the fourier series for f(x)={0,pi -pi<x<0
0<x<pi

This diff. eqn is of the form a \frac{d^3x}{dt^3} + b \frac{d^2x}{dt^2} + c \frac{dx}{dt} + d = f(t) where a,b,c and d are numeric constants. The solution to this is the same as that for 2nd order linear diff. eqns with constant coefficients, provided that f(t) = 0.
So we have \frac{d^3x}{dt^3} - 2\frac{d^2x}{dt^2} + \frac{dx}{dt} = 0 .
The 'kernel' or characteristic equation is in fact:
r³ - 2r² + r = 0
r(r²-2r+1) = 0
r(r-1)² = 0
r= 0, 1 (repeated)
The general solution will be x = (c_1t + c_2)e^t + c_3 where c1, c2 and c3 are constants of integration.

Correct me if i'm wrong as i'm still new to differential equations :)

[roam]

Watch this DE video (http://www.youtube.com/watch?v=Ee45Rn8Qyac). Let me know what you think!

[muhammadasad]

Hello Sir,
I m a student of high energy particle physics. Sir i need the solution manual of Differential equations by S.Balachandra Rao. S.B.Rao is the professor in a college of Banglore, India. Sir please if u can do me a favor, plz give me the solution manual of this book. I shall be very very thankful to u. U can email me on this id lost_somewhere@live.com.

Thank U.....

[Madou]

Where you presented

\frac {d} {dt}[\mu (t)y] = \mu (t) g(t)
I was thrown for a bit.


What does

\frac {d} {dt}[\mu (t)y] = \mu (t) g(t)

mean ? You see I don't have a copy of B&D. )

[bobmerhebi]

hello, I'm also taking a class on ODE but i have a problem -i use An Intro course in Diff. eq.'s by Zill - that i get a nonsense result here is the eq:

sin3x + 2y(cos3x)^3 = 0 (here ^ is to raise a power.how are u raising powers?)

the last result i get which is nonsense ofcourse is: y^2 = -1/6(cos3x)^2. another result includes tan3x but is still negative.

so y^2 is negative which is impossible. is the result right? I think there's a problem with the D.E. given.

hope u can help. thx