Differential equations - Derivations

Discussion in 'Math' started by Skeebopstop, May 31, 2009.

1. Skeebopstop Thread Starter Active Member

Jan 9, 2009
358
3
Hi all,

I've been brushing up again on my differential equations. In reading through the derivations for the use and application of them to model physical systems I have hit a road bump in my intuitive understanding.

The 'general' form of a first order system, after some rearranging to put focus on the time constant, is:

(1/Tau)*dy(t)/dt + y(t) = f(t), where f(t) is some forcing function.

Now if we set f(t) = 0 and provide some y(0) condition, we can solve the 'homogenous' form and get an exponential decay (assuming it is stable).

If we now bring f(t) back in and assume it is not zero, my text states the following for the 'general solution':

y(t) = yh(t) + yp(t), where yh(t) is the exponentially decaying homogeneous form mentioned above, an yp(t) is a 'particular' solution to the forcing function.

This isn't obvious or intuitive to me. Can anyone help me understand how the general form can be proven to provide the general solution?

Regards,

James

2. Mark44 Well-Known Member

Nov 26, 2007
626
1
Linear differential equations like this one come in two flavors: homogeneous and nonhomogeneous.

Homogeneous example: y' + y = 0
Nonhomogeneous related example: y' + y = t

There are many solutions to the homogeneous equation above-- y = ke^(-t) represents all of them. If you have an initial condition given (e.g., y(0) = y_0), you can solve for the constant k.

To solve the related nonhomogeneous equation for a particular solution, the usual technique for the type of equation I showed (linear, constant coefficient) is the method of undetermined coefficients, in which you assume that a particular solution is y(t) = A + Bt. To find the parameters A and B, calculate y'(t), and then y'(t) + y(t) using your supposed solution, and then determine what A and B have to be so that you end up with an equation that is identically true. For my example equation, I get y = -1 + t as the particular solution.

The general solution is made up of two parts: the solution to the homogeneous problem (call it y_h), and the particular solution (call it y_p).

For my made-up problem, the general solution is y(t) = y_h + y_p = ke^(-t) -1 + t.

No matter what k is, the exponential part of the solution will result in 0 when you calculate y' + y (because this exponential function is the solution to the homogeneous problem; i.e., y' + y = 0). This part of the solution also represents the transient behavior of the solution, since ke^(-t) approaches zero as t gets large.

There is an awful lot that I am not saying, that involves concepts from linear algebra such as the kernel of a linear transformation, and a basis for the kernel, and other ideas. And it's difficult to present all of the ideas that are involved in solving differential equations in a small space in a forum reply. If there's anything I said that isn't as clear as you need, let me know.

3. Skeebopstop Thread Starter Active Member

Jan 9, 2009
358
3
Thanks for your reply. Why is it so obvious that the general solution is composed of these two solutions?

I do understand how the homogeneous and particular solutions are found.

Perhaps I have a flaw in my understanding of what a 'particular' solution represents? I have looked it up but still do not get much further in connecting the aforementioned.

Kind Regards,

James

4. Mark44 Well-Known Member

Nov 26, 2007
626
1
A particular solution is a solution to the nonhomogeneous diff. equation. For the example I gave, y_p = -1 + t, so y_p' = 1. Then y_p' + y_p = 1 + (-1 + t) = t, so clearly y_p is a solution to the nonhomogeneous diff equation.

The diff. equation could be written as y' + y = 0 + t. The reason I wrote it that way was to make it a little more obvious what you get when you add the derivative of a function to the function when the function is a) the solution to the homogeneous DE, and b) a particular solution.

A certain kind of function is such that if you add together its derivative and itself, you get zero. Another kind of function is such that if you add together its derivative and itself, you get t. What we're looking for in a general solution is the sum of all such functions in the two categories.

I've shown that such functions exist, but what I haven't shown is that the general solution I've come up with (y = Ke^(-t) -1 + t) is the only possible set of solutions (one for each choice of K, which BTW, can be solved if you have an initial condition). That's more involved than I want to attempt here and now, but there is a theorem that proves this.

5. studiot AAC Fanatic!

Nov 9, 2007
5,005
515
Hello Skee, I don't know if you are familiar with the D operator method?

If not it's very easy. It's just a notation for handling differential expressions
so dy/dx = Dy; d$^{2}$y/dx$^{2}$ = D$^{2}$ etc

and so d$^{2}$y/dx$^{2}$ + dy/dy = F(D)y = (D$^{2}$ + D)y

This treatment may be easier for you to digest.

Now we know from elementary diff =ion theory that we have to add an arbitrary constant for every integration performed. ie one for dy/dx two for the second differential and so on.

If we write any (Mark's) inhomogenous equation in the D operator form

F(D)y = f(x)

We can see that there are no differential coefficients in f(x) so it will not introduce any arbitrary constants. they all come from solving the equation

F(D)y = 0

Which is the associated homogenous equation.

Now Let the solution of F(D)y = u(x) and the general solution of the inhomogenous equation be u(x) + v(x) ie let v(x) be whatever we have to add to u(x) to get the general solution

then

F(D)[u+v] = f(x)
F(D)u + F(D)v = f(x)
but F(D)u = o so

F(D)v=f(x)

v = 1/F(D)f(x)

Which is the particular integral. You have to obtain this by fair means or foul - look it up in a table, guess, ask the plumber or whatever.

6. Wendy Moderator

Mar 24, 2008
20,877
2,655
There is a book we called the CRC (Chemical Rubber ??? something) that tries and does a pretty good job of being the ultimate book of mathematics. The calculus section has a lot derivations that can be used to help with larger problems. Why do I mention this? In many schools Elementary Differential Equations is the precursor to Calculus I and II, and in some curriculum's is also Calculus I. Math had been easy for me at that point, but it was when I knew I was in trouble. I could solve most of the problems (enough to pass the course) but I didn't really understand what I was doing. Didn't help my Calculus instructor had been thoroughly burned out by his instructor, who used nothing but electronics (which would have been excellent for me) and refused to use any electronics examples, which left me cold. I squeaked by Calculus I, then changed my major to Electronics Engineering Technology. The math I already had transfered over.

Wish I could remember those courses, their gone now.

7. Tesla23 Senior Member

May 10, 2009
335
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TO answer your original question, if F and G are solutions to the inhomogeneous equation then F-G is a solution to the homogeneous equation. Conversely, if H is a solution to the homogeneous then F+H and G+H are solutions to the inhomogeneous. So if we can find all the solutions to the homogeneous we have all the possible differences between the inhomogeneous solutions. Hence if we have one particular solution to the inhomogeneous equation and all the solutions to the homogeneous equation we have all the solutions.

8. studiot AAC Fanatic!

Nov 9, 2007
5,005
515
Skee, you asked for a derivation or proof. I realise that my derivation was rather sketchy (hurried) but no one else has actually offered a proof, merely statements and some explanation.

I also realise the D operator is a bit of a suprise for someone not used to it, but there is not a lot more to flesh out the proof. I can try to recast it in more familiar terms if you like.

9. Skeebopstop Thread Starter Active Member

Jan 9, 2009
358
3
Thanks for the responses. Give me a day or two to hash them out.

Much like you Bill, I've made it through two classes on differential equations, a class on vector calc etc.., but when you really look back to how much you got out of uni, you realize it probably wasn't nearly as much as you thought.

I came upon a quote some years back that said, "if a scientist can't teach the most complex concepts to the most uneducated of people, he doesn't understand himself".

That is my approach now so I hope I do not frustrate anyone with what appears a trivial subject

10. Mark44 Well-Known Member

Nov 26, 2007
626
1
CRC is Chemical Rubber Co. They put out a reference titled Standard Mathematical Tables for many years (still do?), and I have the 15th ed. from 1967.

I am skeptical of your statement, Bill, that many schools offer Diff. Eqns as a precursor to calculus. It is most often taught as a successor to calculus, which makes sense, because you have to be able to differentiate and integrate in order to solve differential equations. I taught math at a community college for 18 years, so I have more than a passing knowledge of math curriculum.

In my opinion, real-world examples help motivate the study of differential equations, particularly simple LRC circuits and damped/undamped mass-spring systems, both of which can be represented by the same kinds of differential equation. I was always fascinated by the fact that you can model a theoretical mass-spring-damper system by an equivalent LRC circuit, and vice-versa.

Aug 8, 2005
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