Differential equations

[doraemon]

Hello!

I'm looking for a good introduction to differential equations, if possible online.
Especially, I would like a course that demonstrates how to solve second order
differential equations like ay" + by' + c = K.
That was a long time ago, and I don't remember the detail. I only remember
the results.

Thanks for any hint!

Dora.

 

[rahul.6sept]

U can use Mathematica to solve DE. In mathematica there are commands avble like DSolve or NDSolve which gives symbolic solutions to DEs. u can also use Matlab for the same purpose using the inbuilt functions.

rahul

 

[doraemon]

Hello!

Thanks for your reply. I know I can use Mathematica, but that's not the point. I want to understand
the concept, not solve equations. And I don't find anything online that really explains it all.

Dora.

 

[Ratch]

doraemon,

I really think you need a good textbook on elementary differential equations. That should get you up to speed.

Ratch

 

[_Eduardo_]

Each class of differential equation has its own trick. This an "Homogeneous equation with constant coefficients" --> just google

 

[doraemon]

Hello Eduardo!

Each class of differential equation has its own trick.

The problem is that I am not looking for tricks to solve equations, no cooking recipes,
but good references explaining why these tricks. I remember (more or less) the methods
for solving Bernouilli, Ricatti, Clairaut, Lagrange, ... - type equations, but my problem
is more understanding the thinking process that lead the above people to solve these
equations. I looked at a library in the neighborhood with no more success.

All the online stuff says, basically, that when you have an equation like ay" + by' + c = 0,
you just have to solve the equation ax^2 + bc + c = 0 and it gives you 2 roots r1 and r2 and
therefore the solutions of the differential equation are c1 * exp(r1t) and c2 exp(r2t).

Now since I don't like the above "therefore" because it does not explain anything, I searched further
and found out another pages saying that the strategy to solve that kind of equation is to look
for solutions like exp(rt). Indeed, if y = exp(rt), then y' = r*exp(rt) and y" = r^2*exp(rt).
I get therefore ar^2*exp(rt) + br exp(rt) + cexp(rt) = 0, and since exp(rt) is never 0,
it can be pulled out and ar^2 + br +c is therefore 0, producing 2 roots (or one in some cases),
and these roots indicate damping / amplification, or oscillation depending whether they are
real or complex, which makes perfectly sense to me.

At that point I am just asking myself: why is this THE strategy. Why isn't there another
strategy using another weird function instead of the well known exponential?

Thanks for any hint.

Dora.

 

[albbg]

try googling "differential equations" in google books. You can preview the books on that topic and then buy the more useful from your point of view.

 

[Ratch]

doraemon,

I remember (more or less) the methods
for solving Bernouilli, Ricatti, Clairaut, Lagrange, ... - type equations, but my problem
is more understanding the thinking process that lead the above people to solve these
equations.

Those mathematicians have been dead for many years now, so no one can ask them how they came up with their methods of solution. All you can do is look at the proofs of their solution and admire their perspicacity. A good textbook on the subject will present and elucidate those solution methods.

At that point I am just asking myself: why is this THE strategy. Why isn't there another
strategy using another weird function instead of the well known exponential?

'Strategy' is a poor choice of words. Solving an equation is an application consisting of recognizing the type of equation and applying the solution method. No strategy involved.

Ratch

 

[_Eduardo_]

At that point I am just asking myself: why is this THE strategy. Why isn't there another
strategy using another weird function instead of the well known exponential?

Since the general solution of this equation is A + B e ^ (pt) + C e ^ (qt), the simplest method to solve will be replaced in the equation and finding the coefficients.

But... Why is this the general form?
That's part of the study of DE, which is not limited to finding "a useful function to replace", but also to see the existence and how many of linearly independent solutions. As I remember, this was done by matrices.

 

[pcsandhya82]

Hi,

Online books you have to google or find in rapidshare. Else if you go for actual books, Kreyzig is a good option. Infact it is available online somewhere.

 

[ahsan_i_h]

hello doraemon,

Lets discuss about differential equation
There are two methods:
a)Trial solution method (which you did not like)
b)succesive integration method (It is a real systematic but has a problem of understanding D operator. I tried many days to understand D operator, then came to a conclusition)

Now do you understand D operator? did you hear anything about D operator?

 

[doraemon]

Hello!

a. The point is not that I don't like the trial solution. I like any solution if it's rigorously proved.
But all demonstrations I have seen on the net start with "the solution is A + B e ^ (pt) + C e ^ (qt)".
Proving that the above function is one solution (or one family of solutions) is fine.
However, how do I prove that it is the only family of solutions.
Or in other words, if I suppose that there is another class of functions which is not a linear
combination of e^kt, then how do I prove that this class cannot exist?

b. What do you call D operator? Is it the Laplace operator (also sometimes called s or p)?

Dora.

 

[Ratch]

doraemon,

However, how do I prove that it is the only family of solutions.
Or in other words, if I suppose that there is another class of functions which is not a linear
combination of e^kt, then how do I prove that this class cannot exist?

You cannot have a solution of a linear differential equation be a nonlinear equation. The derivative of a nonlinear equation will not be linear, so it cannot possibly be a solution to a linear differential equation.

What do you call D operator? Is it the Laplace operator (also sometimes called s or p)?

No, it is not the LaPlace operator. It is simply a shortcut for d(f(x))/dx. For instance, D(x²+x) = 2x+1

Ratch

 

[ahsan_i_h]

Hello, Dora , I am reviewing my differential equations. I will join you soon.

 

[rahul.6sept]

Hi Ratch,

Can u plz elaborate regarding the technique for solving a non-linear differential equation?

Regards,
rc

 

[Ratch]

rahul.6sept,

Can u plz elaborate regarding the technique for solving a non-linear differential equation?

It all depends of the form of the differential equation. Is it exact? Does it have an integrating factor? Are the variables separable? Is it homogeneous? Can a variable be substituted for an expression. I cannot give you a course on solving differential equations. That is what a good textbook is for.

Ratch

 

[enob]

This guy is pretty good. I used him a lot when I was taking a class on it.

https://www.youtube.com/playlist?list=PLD4B0062CA82D73FB&feature=plcp

 

[ahsan_i_h]

hello dora,

differential equations like ay" + by' + c = K.

I think there must be cy in place of c.

Now for the first order linear differential equation with constant coefficient, like
y' - a0y=g(x) we know the solution is,
y=e^a0x [ integration sign ] e^-a0x g(x) dx (Put integration constant after integration)

Now lets try to solve second order linear differential equation with constant coefficient,
WITHOUT assuming any type of format of the solution at begining.

The differential equation to be soved is
y'' + a1y' + a0y=g(x)

Now we can think that this second order differential equation is formed by two first order differential equation like,
(y' - r0y)' - r1(y' - r0y)=g(x) .......... The idea is, one first order equation inside another first order equation.
Now we have to find the values of r0 and r1.

By equating the coefficients of above two equations we get
a1= - (r1+r0) and a0= r1r0. Now putting r1=a0/r0 in a1= - (r1+r0) we get,
r0^2 + a1r0 + a0=0 ........... This is the so called characteristic equation.
So by solving r0^2 + a1r0 + a0=0 we will get the values of r1 and r0.
Now we have the values of r1 and r0

So lets start to solve the equation,
(y' - r0y)' - r1(y' - r0y)=g(x)
Now, (y'-r0y) = e^r1x [ integration sign ] e^-r1x g(x) dx .......We got this, using formula of first order.
So, y = e^r0x [ integration sign ] e^-r0x e^r1x [ integration sign ] e^-r1x g(x) dx dx .......We got this same way.

Summing up everything we get,
The second order linear differential equation with constant coefficient like,
y'' + a1y' + a0y=g(x)
The solution is,
y = e^r0x [ integration sign ] e^-r0x e^r1x [ integration sign ] e^-r1x g(x) dx dx (Put integration constant after each integration)
Where r0 and r1 are the roots of [anything]^2 + a1[anything] + a0 = 0.
NOTE: If you do not put integration constant after each integration, then you will get only particular integral (means steady state solution)
Integration constants are responsible to make complementary function (means transient solution)

Last of all,
Sorry about the way I asked about D operator. Certainly you know that D is d/dx.
In text books(as far I remember "analysis of linear systems" by D.K Cheng) they say that D operator follows some basic rules.
These basic rules are the basic rules behind multiplication symbol or addition symbol or parentheses symbols.
So they solve first order equation,
y' - a0y=g(x) by writing it like,
(D - a0)y=g(x)
So y= [1/(D-a0)] g(x)
But it is true that the solution is,
y=e^a0x [ integration sign ] e^-a0x g(x) dx
So they DEFINED [1/(D-a0)]g(x) AS e^a0x [ integration sign ] e^-a0x g(x) dx
Now for second order differential equation,
y'' + a1y' + a0y=g(x) they write the equation like,
(D^2 + a1D + a0) y=g(x) so they start to factoring the left side like,
(D - r1)(D - r0)y=g(x) then,
(D - r0)y=[1/(D - r1)]g(x) then,
(D - r0)y=e^r1x [ integration sign ] e^-r1x g(x) dx ....... Using the DEFINATION of [1/(D - a0)]g(x). then,
y=[1/(D - r0)] e^r1x [ integration sign ] e^-r1x g(x) dx then,
y= e^r0x [ integration sign ] e^-r0x e^r1x [ integration sign ] e^-r1x g(x) dx dx .....Same way using the DEFINATION.
I could not like this freedom of D operator(may be some day I will like it), thats why I tried many days to understand D operator.
Then I came to a solution without D operator which I already shown.

 

[rahul.6sept]

Hi,

Someone plz let me know the procedure for normalising a differential equation. I need it but don't know as how to proceed for the same.


Regards,

Rahul

 

[milind.a.kulkarni]

Hi Rahul

can you post an example here what kind of differential equation that you are taking about....

with regards,