Intro. to Differential Equations

Tom Mattson

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.

[tex] xy^2 - e^{-y} - 1 = 0 [/tex]

[tex] u = x, v = y [/tex]

[tex] (u + du)(v+dv)^2 = (u + du)(v^2 + 2vdv + d^2v) [/tex]

[tex] = uv^2 + 2uvdv + ud^2v + v^2du + 2vdvdu + d^2vdu [/tex]

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

[tex] 2uvdv + v^2du [/tex]

We then plug in for x and y and get

[tex] 2xyy' + y^2 [/tex]

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

[tex] z = -y [/tex]

[tex] -e^{-y} = -e^z [/tex]

[tex] dz = -dy [/tex]

[tex] \frac {d(-e^z)} {dz} = -e^z = -e^{-y} [/tex]

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

[tex](dz)(-e^{-y}) = + e^{-y}y' [/tex]

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

[tex] 2xyy' + e^{-y}y' + y^2 = 0 [/tex]

As implicit differentiation of [tex] xy^2 - e^{-y} - 1 = 0 [/tex]

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

[tex] (xy^2 + 2xy - 1)y' + y^2 = 0 [/tex]

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

[tex] e^{-y} = xy^2 - 1 [/tex]

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

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

Now this is where I need help. Can I say that the original equation [tex] xy^2 - e^{-y} - 1 = 0 [/tex] is an implicit solution to [tex] (xy^2 + 2xy -1)y' + y^2 [/tex], 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 [tex] xy^2 - e^{-y} - 1 = 0 [/tex] is an implicit solution to [tex] (xy^2 + 2xy -1)y' + y^2 [/tex]

Fritz

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

QuantumTheory

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, [tex]dx^2/dy^2[/tex], 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 [tex]dx[/tex]. So if the derivative is the rate of some curve(therefore the height of the curve), lets call it [tex]f(t)[/tex].

Then,

[tex]f(t)[/tex] = [tex]f(t)dx[/tex]


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:

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

dextercioby

Which one,the first statement?

Daniel.

whozum

How does:

follow from

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

Just use _ and ^ in LAtex its pretty easy

Ruslan_Sharipov

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:
[tex]
y''=P(x,y)+3 Q(x,y) y'+3 R(x,y) (y')^2+S(x,y) (y')^3,
[/tex]
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:
[tex]
\tilde x=\tilde x(x,y),
[/tex]
[tex]
\tilde y=\tilde y(x,y).
[/tex]
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

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:

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

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:
[tex]f(x)= \frac {g}{h}[/tex]

[tex]f'(x)=\frac {g'h-h'g}{h^2}[/tex]


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.