General solution vs particular solution

  • #36
Mark44 said:
With regard to ##y'' + 9y = \frac 1 2 \cos(3x)##:Sure, this is a particular solution, one that can be decomposed into ##\frac x {12}\sin(3x)## and ##\frac 1{12}\sin(3x)##. The former is a solution to the nonhomogeneous DE above, and the latter is a solution of the homogeneous equation.
My whole argument here is why go out of one's way to devise a particular solution that isn't disjoint from the homogeneous problem's solution set?
Define "disjoint" in this context.
 
Physics news on Phys.org
  • #37
I have read all of your comments.
I will post an image from my textbook, and would like all of your opinion on it.
1710259398910.png

Part a) is fine, just differentiate the top expression with respect to t and make the appropriate substitutions. Part b) asks you to find a general solution, so we find a particular solution (namely ##x(t)=10##) and add that onto the general solution of the inhomogeneous equation. Part c) is also straightforward (use the top equation to find ##y(t)## by differentiating ##x(t)## and making the appropriate substitution). However in part d) we are given some initial conditions and then asked to find the particular solution. Technically speaking, are the functions we found (such as ##x(t) = 10##) not also particular solutions? Clearly what makes the particular solution in part d) different to ##x(t) = 10## is that it matches the required initial conditions.
Thoughts?
My textbook calls the function ##x(t) = 10## the particular integral, and reserves the phrase particular solution for the final answer with no undetermined constants. Found a similar post, will link below. https://www.physicsforums.com/threa...ular-integral-and-particular-solution.667568/
 
  • #38
sdfsfasdfasf said:
I have read all of your comments.
I will post an image from my textbook, and would like all of your opinion on it.
View attachment 341670
Part a) is fine, just differentiate the top expression with respect to t and make the appropriate substitutions. Part b) asks you to find a general solution, so we find a particular solution (namely ##x(t)=10##) and add that onto the general solution of the inhomogeneous equation. Part c) is also straightforward (use the top equation to find ##y(t)## by differentiating ##x(t)## and making the appropriate substitution). However in part d) we are given some initial conditions and then asked to find the particular solution. Technically speaking, are the functions we found (such as ##x(t) = 10##) not also particular solutions? Clearly what makes the particular solution in part d) different to ##x(t) = 10## is that it matches the required initial conditions.
Thoughts?
This needs to be in a new thread, IMO.
 
  • #39
PeroK said:
This needs to be in a new thread, IMO.
It is still under the main topic question "What actually is a particular solution?".
 
  • #40
Mark44 said:
It's relevant to me in the way I approach these kinds of problems; namely, I find a basis of functions that solve the homogeneous problem, and then find a particular solution that is disjoint from that basic set.
Any function that solves the non-homogeneous equation is a particular solution and therefore is explicitly disjoint from the homogeneous solutions. As we have discussed and shown, there is no uniqueness here. Any particular solution will do.

This is very standard and a basic fact about linear differential equations. You simply cannot make the distinction you are attempting to do in singling out a specific particular solution. As @PeroK said, you would have to mathematically be able to define the properties of such a unique function. ”Does not contain a part that is on the homogeneous form” will not do as previously shown as functions may be rewritten on different forms where a homogeneous solution term pops out. So I ask you again to provide your actual mathematical definition.

For the solution process, it does not matter which particular solution you choose, but teaching that there is only one unique particular solution borders on spreading misinformation in my book.

Mark44 said:
that is disjoint from that basic set.
Again, this is not well defined. I know what you are trying to say, but things simply do not work like that and unless you specify the mathematical properties of the function you want to choose, it will remain ambiguous. There is also nothing stopping you from adding an arbitrary fixed homogeneous solution to a particular solution and get a new particular solution that you can work with just as well. In fact, the preferable way to solve these problems should be to first find the particular solution, which reduces the remaining problem to solving the homogeneous part, which is more transparent in my opinion.

Mark44 said:
With trig or hyperbolic trig functions, one can find other functions that are identically equal. The example you gave of 1=2cos2⁡(x)−cos⁡(2x) falls into this category. Can you argue that the two sides of this equation are actually different functions?
Obviously they are the same function, but you are missing the point. The point is that 1 can be rewritten in a manner such that it is not disjoint from the homogeneous (your words).
 
  • Like
Likes docnet and PeroK
  • #41
sdfsfasdfasf said:
However in part d) we are given some initial conditions and then asked to find the particular solution.
No. You are asked to find the particular solution that satisfies <<conditions>>. Here the <<conditions>> (typically boundary/initial conditions) single out one of the many particular solutions.
 
  • #42
PeroK said:
Define "disjoint" in this context.
-- is not a sum that contains at least one function that is a linear combination of the homogeneous equation's basis set.
 
  • #43
Orodruin said:
In fact, the preferable way to solve these problems should be to first find the particular solution, which reduces the remaining problem to solving the homogeneous part, which is more transparent in my opinion.
I'm glad you added the "in my opinion part." The way I was taught and later taught this subject, was to find the solution to the homogeneous problem, and then find a particular solution.
One technique that I found to be fruitful was the so-called method of annihilators.
Using your example again, y'' + 4y = 4, this can be written as ##(D^2 + 4)y = 4##.
Since the D (or differentiation) operator annihilates any constant the nonhomogeneous problem above can be rewritten as a higher order homogeneous equation as ##(D^3 + 4D)y = D(4) = 0##.
Factoring the left side gives ##D(D + 2i)(D - 2i)y = 0##, so a basic set for this new equation is ##\{1, e^{2ix}, e^{-2ix}\}##. De Moivre's formula allows us to change this set to ##\{1, \cos(2x), \sin(2x)\}##. Of course the latter two functions are in the kernel of the homogeneous equation, so I see little point in including them as part of a particular solution.
A particular solution can be found by determining which multiple of the basis function 1 satisfies the nonhomogeneous equation. Letting ##y_p = A \cdot 1##, we have ##y_p'' + 4 y_p = 4##. Since ##y_p'' = 0##, we have ##4y_p = 4 \Rightarrow y_p = 1##.
This leads us to a general solution of the original nonhomogeneous ODE being ##y_g = c_1\cos(2x) + c_2\sin(2x) + 1##, where the two constants can be determined if initial conditions are included.
 
  • #44
Mark44 said:
-- is not a sum that contains at least one function that is a linear combination of the homogeneous equation's basis set.
But, as has been shown with several examples, this does not uniquely identify one of the particular solutions. It is therefore not a mathematically rigorous way of singling out ome of the many particular solutions.

More to the point: There is absolutely nothing wrong with using a particular solution that has a term that is a homogeneous solution. You will get different constants when you adapt the general solution to whatever initial/boundary conditions you may have, but the final solution will be the same.

This is where teaching that there is only one permissible particular solution goes wrong. What is worse, you might mark down someone who used ##\sinh(x)## instead of ##\cosh(x)## or ##e^x/2## as their particular solution when in reality they did something completely fine.

One should also note that a particular solution that also satisfies the initial conditions is what is required to solve a diffential equation. It is explicitly what the OP’s problem asks for.
 
  • #45
Orodruin said:
More to the point: There is absolutely nothing wrong with using a particular solution that has a term that is a homogeneous solution.
Other than it is a waste of effort? The term that is a homogeneous solution vanishes when it is substituted in the DE. That's really my main point.

Orodruin said:
You will get different constants when you adapt the general solution to whatever initial/boundary conditions you may have, but the final solution will be the same.
If we both arrive at the same solution, then what exactly is the problem?

Orodruin said:
This is where teaching that there is only one permissible particular solution goes wrong. What is worse, you might mark down someone who used ##\sinh(x)## instead of ##\cosh(x)## or ##e^x/2## as their particular solution when in reality they did something completely fine.
This is something I strove hard never to do. When I was marking homework or exams, and someone had an answer that looked different from the answer sheet I had prepared, but was equivalent to mine, I never marked them down. Never.
 
  • #46
Mark44 said:
Other than it is a waste of effort? The term that is a homogeneous solution vanishes when it is substituted in the DE. That's really my main point.
There really is no point here. As has been shown repeatedly and with different examples, it is not always directly obvious that two different homogeneous solutions have terms from the homogeneous solution.

They must of course differ by a homogeneous solution, but they are always equally permissible and teaching that there is only a single particular solution is simply wrong. All solutions where you fix the arbitrary constants are particular solutions by definition. That you choose to use a specific one out of those is a matter of your preferred way to solve a problem, not one of making a particular solution unique

Mark44 said:
This is something I strove hard never to do. When I was marking homework or exams, and someone had an answer that looked different from the answer sheet I had prepared, but was equivalent to mine, I never marked them down. Never.
What if they only solved the problem partially and used a different particular solution? The final answer is not everything.

Mark44 said:
If we both arrive at the same solution, then what exactly is the problem?
Nothing. It is you who seems to have a problem with there being more than one particular solution.
 
  • #47
Mark44 said:
-- is not a sum that contains at least one function that is a linear combination of the homogeneous equation's basis set.
Okay, so if we have explicit functions, you can look at the function and check that it meets this criteria. But, what if we have a particular solution in theory: ##y_p(x)##. How do we confirm that we have the unique particular solution? Suppose you say you have found the unique particular solution. How can you prove it?

I would like to see your proof that ##y_p(x)## is not a sum that contains at least one function that is a linear combination of the homogeneous equation's basis set. I.e. a proof that ##y_p(x)## can be determined uniquely?
 
  • #48
PeroK said:
I would like to see your proof that ##y_p(x)## is not a sum that contains at least one function that is a linear combination of the homogeneous equation's basis set. I.e. a proof that ##y_p(x)## can be determined uniquely?
And I would like to see a valid method for squaring the circle while we are at it ... 😏
 
  • #49
Orodruin said:
As has been shown repeatedly and with different examples, it is not always directly obvious that two different homogeneous solutions have terms from the homogeneous solution.
I think you got in a hurry here and wrote something different from what you meant.

Orodruin said:
They must of course differ by a homogeneous solution, but they are always equally permissible...
A point I have repeatedly tried to make.

Orodruin said:
... and teaching that there is only a single particular solution is simply wrong.
Okay, okay, please stop beating this dead horse! Our only disagreement here seems to be that my personal definition of the term "particular solution" is more restrictive than yours, making it at odds with the prevalent definition. So mea culpa. In my usual approach to these kinds of problems, I look for the basis functions for the homogeneous problem, and then look for a function that solves the nonhomogeneous problem, one that does not contain terms that are any linear combination of the basis functions for the solution of the homogeneous problem.

Regarding the examples given so far:
##y'' + 9y = \frac 1 2 \cos(2x)##
Homogeneous problem basis set: ##\{\sin(3x), \cos(3x) \}##
For the nonhomogeneous problem I would consider multiples of ##\{x\sin(3x), x\cos(3x) \}##
I will admit that ##\frac{x + 1}{12}\sin(3x)## is a particular solution, but not one that I would choose, as it can be decomposed into a sum with a term from the basis set
(##\frac 1{12}\sin(3x)##) for the homogeneous problem. It's also more complicated than what is absolutely necessary.

##y'' + 4y = 4##
Homogeneous problem basis set: ##\{\sin(2x), \cos(2x) \}##
For the nonhomogeneous problem I would consider a constant function; i.e., a multiple of 1.
I will admit that ##2\cos^2(x)## is a particular solution, but not one I would choose. This can be decomposed into ##\cos(2x) + 1##, which contains a term from the homogeneous problem basis set. It too is also more complicated than what is absolutely necessary.
I also would not consider any function that is identically equal to some constant, such as ##\sin^2(kx) + \cos^2(kx)##, ##\sec^2(kx) - \tan^2(kx)##, etc.

##y' + y = e^x##
Homogeneous problem basis set: ##\{e^{-x}\}##
I will admit that ##\sinh(x)## and ##\cosh(x)## are potentially particular solutions, but not ones that I would choose over ##e^x## as they would include a term from the homogeneous problem basis set. These hyperbolic trig functions are also more complicated than what is absolutely necessary.
 
  • #51
I think that particular solutions may come in hand when you already know the basics and just shorten the solution, they are more of shortcuts and I am also using these from time to time. I didn't want to continue this topic I just wanted to express my idea, sorry if it got you angry, also nice to meet you all.
 
  • #52
Mark44 said:
A point I have repeatedly tried to make.
Well, your starting point in this thread was that there was one and only one particular solution and no other solution could possibly be called a particular solution. This is what we are claiming to be false and arguing against. This
Mark44 said:
The particular solution is what you (@sdfsfasdfasf) showed, namely, yp(x)=112sin⁡(3x). No other function will satisfy this ODE.
is false and is what started this whole mess. (My emphasis)

Mark44 said:
Our only disagreement here seems to be that my personal definition of the term "particular solution" is more restrictive than yours, making it at odds with the prevalent definition.
(My emphasis)

If you use your personal definition in teaching, your students will inherit the same disconnection from standard terminology. I believe that this is the issue here.

Mark44 said:
I look for the basis functions for the homogeneous problem, and then look for a function that solves the nonhomogeneous problem, one that does not contain terms that are any linear combination of the basis functions for the solution of the homogeneous problem.
That's fine of course, but the point is that the method of using a particular solution (any particular solution!) to homogenise the differential equation is not restricted to this way of doing things. You are free to choose any particular solution you can get your hands on, but you have no basis in claiming that it is a preferred particular solution over any other.

Mark44 said:
These hyperbolic trig functions are also more complicated than what is absolutely necessary.
I could argue that they are about half as complicated as the exponential function. Their definitions in terms of series expansions contain only odd/even terms!
Mark44 said:
This can be decomposed into cos⁡(2x)+1, which contains a term from the homogeneous problem basis set.
I do not agree with this argument as it can just as well be applied to the function 1 as well. You can rewrite 1 as a sum which contains a cos(2x) apart from the squared cosine. I am not saying it looks more beautiful, but which function is the other plus a homogeneous part is reflexive.
Mark44 said:
but not ones that I would choose over ex as they would include a term from the homogeneous problem basis set.
But you can just as well write that ##e^{x}/2 = \cosh(x) - e^{-x}/2## so ##e^x## includes a term from the homogeneous problem basis set! This simply is not an argument for one over the other. If you have ##f_1(x)## and ##f_2(x)##, which are particular solutions such that ##f_2(x) = f_1(x) + h(x)##, where ##h(x)## is a homogeneous solution, then ##f_1(x) = f_2(x) - h(x)##, where ##-h(x)## is a homogeneous solution. Do not be fooled by the fact that you may be more familiar with ##f_1(x)## than you are with ##f_2(x)##!
 
  • #53
This thread did well, I am happy.
 
  • Like
Likes JamalGross, docnet, SammyS and 1 other person

Similar threads

  • Differential Equations
Replies
1
Views
736
  • Differential Equations
Replies
1
Views
1K
Replies
3
Views
3K
  • Differential Equations
Replies
8
Views
2K
  • Differential Equations
Replies
5
Views
720
  • Differential Equations
Replies
2
Views
1K
  • Differential Equations
Replies
7
Views
465
Replies
1
Views
2K
  • Differential Equations
Replies
5
Views
1K
  • Differential Equations
Replies
8
Views
2K
Back
Top