differential equation problem

https://www.amazon.co.uk/Mathematics-Electrical-Technicians-Longman-Technician/dp/0582234212

This used to be our bible at college. It has several chapters on solving such equations.

WTP Pepper said:

https://www.amazon.co.uk/Mathematics-Electrical-Technicians-Longman-Technician/dp/0582234212

This used to be our bible at college. It has several chapters on solving such equations.

Click to expand...

Are you sure? That book does not appear to be at the level required for this problem. What are the chapter titles that you think apply? Can you scan any relevant pages and post them?

The reason that I'm being cautious here is that someone (in particular, the OP Georg) may buy this book based on your recommendation that it will help solve this problem, and it could be a waste of money for him.

Hi,


Yes i had my doubts too for that book based on the description, but it's hard to tell. Perhaps a few snapshots of the guts would reveal more info about what it contains.

How about "Differential Equations for Dummies" ha ha. Seriously though, this equation is 'different' because it has one of the derivatives squared. That's not ordinary. Usually we see an equation that depends on y and it's derivatives: F(x,y,y',y", ...) but not on one or more of it's derivatives squared: F(x,y,(y')^2,y"). I think that makes it "not ordinary" in a manner of speaking.

If I wasn't sure I wouldn't waste my time posting.

I'm not prepared to start posting photocopied texts from copyrighted material here. It's levelled at final year degree engineering mathematics in the UK (or as it was in 1987). That's on a par to a post grad Doctorate to you in the US

However...
Chapter 20
The solution of linear second order differential equations of the form a(d2y/dx2)+b(dy/dx)+cy=0

Chapter 21
The solution of linear second order differential equations of the form a(d2y/dx2)+b(dy/dx)+cy=f(x)

In Chapter 20, make b=0 and you have the same equation the OP was looking at.

Problem 4 d2y/dx2-(y*n^2)=0 sounds pretty close
Problem 20 d2x/dt2+225x=0 gives x=cos15t

"Differential Equations for Dummies". Not here in the UK.

WTP Pepper said:

If I wasn't sure I wouldn't waste my time posting.

I'm not prepared to start posting photocopied texts from copyrighted material here. It's levelled at final year degree engineering mathematics in the UK (or as it was in 1987). That's on a par to a post grad Doctorate to you in the US

However...
Chapter 20
The solution of linear second order differential equations of the form a(d2y/dx2)+b(dy/dx)+cy=0

Chapter 21
The solution of linear second order differential equations of the form a(d2y/dx2)+b(dy/dx)+cy=f(x)

In Chapter 20, make b=0 and you have the same equation the OP was looking at.

Problem 4 d2y/dx2-(y*n^2)=0 sounds pretty close
Problem 20 d2x/dt2+225x=0 gives x=cos15t

"Differential Equations for Dummies". Not here in the UK.

Click to expand...

Hi,


Well the question was never whether you were sure or not, the question was whether or not the book applied to this problem. From what you are now stating it appears that our guesses were correct: that it does not apply.

It seems that this book does not apply because the two chapters you are quoting are for linear differential equations and this one is not linear. And there is usually nothing "close to" that which will work unfortunately, it has to be the same or at least contain the same derivatives exactly as they appear in the original equation or an alternate form.

In other words, it has to contain the second derivative of y^2 and the function y iself, or else the function itself y and it's second derivative and also it's first derivative as your examples might show, but that first derivative has to be squared not standing alone even though maybe with a coefficient like K*y'.

So we have to see the equation F with dependencies like this:
F(y,(y^2)")
or
F(y,(y')^2,y")

so if you like you can look for an equation with y and (y^2)", or an equation with y, (y')^2, and y" in it. At this point i would be happy to see it even without coefficients.

Thanks.

Also, since when does a foreign degree increase when migrating to the US ?
Usually it is harder to get approved here

WTP Pepper,

I agree with MrAl. The book in question does not address the particular complexities of this problem we are talking about. I'm sure it is a good book, and well worth the money someone spends on it. It's just that we need to be clear why we are recommending it so that a person will purchase it for the right reason. That material is very appropriate to prepare someone to approach this problem, but it does not address this type of problem directly.

Also, you said this is at a post-grad level, but my initial concern was that the US version of Amazon said "It will also meet the needs of first and second year undergraduates studying electrical engineering." That is quite a different statement, so I'm confused as to what level this text is at.

Also, on the issue of copyright, the posting of 1 or 2 pages from a large text book is not in violation of copyright and is allowed under the fair-use clause, given that it is for education and not for profit and would be too small a portion of the book to be useful without actually buying or borrowing the full text. Typically, this type of posting can lead to increased sales of the book, so I really don't think they would mind. Unfortunately, the Amazon website does not have preview for this book, and a view of a couple of pages might even inspire me to buy the book if I like what I see. Also, consider that typically these books are mostly rehashing long-standing methods that either exist in the public domain, or are "taken" from other sources too.

Anyway, it's good you are trying to help and provide more information. We are just trying to be clear on what the exact nature of the help will be.

OK from memory in terms of x & y
y"=d2y/dx2
So does (y^2)"=d2y2/dx2 or d/dx (d/dx (y^2) ) ) ?

Hi guys,


Steve:
Thanks for remembering to quote the fair use of copyrighted material. Although it is not super clear what is fair use and what is not, i think the Author would be grateful if he saw one of his equations quoted here combined with "it's a very good book that i highly recommend" or something like that
I know i would be tempted to buy the book if it did contain that equation.

WTP Pepper:
y" is equivalent to d^2y/dx^2 or:
[LATEX]\frac{{d}^{2}(y)}{d\,{x}^{2}}[/LATEX]

(y^2)" is:
[LATEX]\frac{{d}^{2}(y^2)}{d\,x^2}[/LATEX]

or (y^2)" is:
[LATEX]\frac{d}{dx} (\frac{{d}(y^2)}{d\,{x}})[/LATEX]

Usually we see the equation depend on one or more of these:
x,y,y',y",y"',y"", etc.,

but we are seeing here either these two:
y,(y^2)"

or these three:
y,(y')^2,y"

in the equation.

BTW thanks much for your efforts here. Between the four of us we might figure this out eventually

I have looked in my library of text books from my academic years and find nothing re d/dx(y^2). I am too old to try and do it by hand so I bow out gracefully,

I won't say I'm bowing out here, but I also don't have much else to contribute. The opinion I offered in post #13 is my best attempt to provide an answer. I also tried to continue along the lines of misterT but did not find an closed form solution via that route. However, misterT's approach might lend itself to easier numerical solutions, so I find this an interesting approach. We have to keep in mind that there is no guarantee that closed form solutions exist for the most general case. I'll keep an eye on this thread to see if more progress is made.

Hi,

Sorry to hear that WTP Pepper, but if you are tired of looking for solutions i guess i cant really blame you as this is a different problem from the normal one.

Steve, i was able to do a numerical with two first order coupled equations and that seems to be the easiest to do, but i am open to just about anything at this point Supposedly with the two first order equations we could look at the phase space and determine various things about the equation such as if a solution exists, but i'd really have to read up on this.

All:
Also, i have a feeling what we found means that this equation doesnt have a solution. My reasoning is that if we found a trial solution that actually works in the DE but does not lend itself to a general solution, then there is probably no solution. Or possibly if that solution works in the DE then a mariad of solutions exists so there's no real solution either. I have a feeling this is a theory but im too lazy to read up on this subject of finding out if a solution exits or not. Maybe in the future some time.

I also tried misterT's solution try, taking it step by step as he was illustrating nicely, although i get stopped when it comes to "integrate" because there is no way to integrate y not having the function it represents, without doing it numerically, unless there is a trick of some kind we can use, but i dont know what it is at the moment or if it even exists. If i integrate the way he showed it would have been integrated, that means integrating with respect to y itself which of course just does not work because y is not a variable it is a function as illustrated here:

y=f(x)

then what is Integrate(y,x) equal to? There's no way to know without information about f(x).

Consider a slightly simpler problem:

y"+y'= -y

If we try to integrate directly with respect to y we get something like:
y'+y=-y^2/2

and subtracting y we get:
y'=-y^2/2-y

and integrating we get:
y=-y^3/6-y^2/2-y+K

and solving for y now we get a real constant y=K2 and either two more real or a complex pair.

But the real solution is in the basic form of:
y=e^(a*x)*(k1*sin(W*x)+k2*cos(W*x))

so how do we get those trig functions and exponential from a direct integration? Of course we dont so we cant do it that way