General solution vs particular solution

  • #1
sdfsfasdfasf
57
8
Homework Statement: What actually is the particular solution of an ODE?
Relevant Equations: x

Consider the differential equation ##y'' + 9y = 1/2 cos(3x)##, if we wish to solve this we should first solve the auxiliary equation ##m^2 + 9 = 0## giving us ##m=3i,-3i##, this corresponds to the complementary function ##Asin(3x) + Bcos(3x)##. Then because the complementary function contains the RHS of the differential equation our guess for the particular integral is no longer ##Csin(3x) + Dcos(3x)## but rather ##x(Csin(3x) + Dcos(3x))##. Then we will find the derivatives and substitute into our original equation to find that ##C=1/12## and ##D=0##. Now say we were given some intial conditions and found that ##A=B=1##, is the "particular solution" ##1/12xsin(3x)## or is the particular solution ##sin(3x) + cos(3x) +1/12xsin(3x) ##? My textbook sometimes references the PI + CF with the constants evaluated as the particular solution, whereas other times it references the PI as the particular solution. This makes me confused.
 
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  • #2
There's no such thing as "the" particular solution. There is only "a" particular solution. Any one will do.
 
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  • #3
Alright. So we have infinitely many particular solutions, and only one will satisfy our initial conditions as well as the differential equation? I think my confusion stemmed from the fact that our final answer will be a particular solution, aswell as all the other particular solutions which don't match the initial conditions.
 
  • #4
I'm going to disagree with @PeroK. The original ODE is given without initial conditions. The particular solution is what you (@sdfsfasdfasf) showed, namely, ##y_p(x) = \frac 1{12}\sin(3x)##. No other function will satisfy this ODE.
The general solution would be ##y(x) = c_1\sin(3x) + c_2\cos(3x) + \frac 1{12}\sin(3x)##. Initial conditions, if given, would enable you to determine ##c_1## and ##c_2##.
 
  • #5
sdfsfasdfasf said:
Alright. So we have infinitely many particular solutions, and only one will satisfy our initial conditions as well as the differential equation?
There are infinitely many particular solutions. The idea is that you only need to find one. Any one will do. Then any solution can be expressed as that particular solution (I guess at this point, you could call it "the" particular solution you've chosen) and a solution to the homogeneous equation. That's because the difference of two solutions is a solution to the homogeneous equation.
sdfsfasdfasf said:
I think my confusion stemmed from the fact that our final answer will be a particular solution, aswell as all the other particular solutions which don't match the initial conditions.
Satisfying initial conditions is something else. According to this link, this is called the "actual" solution(!)

https://tutorial.math.lamar.edu/classes/de/definitions.aspx

By the way, that website is a great resource for all things calculus.
 
  • #6
Mark44 said:
I'm going to disagree with @PeroK. The original ODE is given without initial conditions. The particular solution is what you (@sdfsfasdfasf) showed, namely, ##y_p(x) = \frac 1{12}\sin(3x)##. No other function will satisfy this ODE.
Yes, there are other functions. ##y_p(x) + y_h(x)##, where ##y_h(x)## is any solution to the homogeneous equation is also a particular solution.
 
  • #7
PeroK said:
Yes, there are other functions. ##y_p(x) + y_h(x)##, where ##y_h(x)## is any solution to the homegeneous equation is also a particular solution.
I was distinguishing between the particular solution, ##y_p## and the doubly-infinite set of general solutions, all of which include the particular solution.
 
  • #8
Mark44 said:
I was distinguishing between the particular solution, ##y_p## and the doubly-infinite set of general solutions, all of which include the particular solution.
That's a false distinction. There's no way to say that ##y_p## doesn't have any of the homogeneous solutions in it. How do you identify mathematically the one "true" particular solution?

In practice, of course, it hardly matters. You can pretend there is only one allowable particular solution. But, that assertion doesn't stand up to mathematical scrutiny.
 
  • #9
PeroK said:
That's a false distinction. There's no way to say that ##y_p## doesn't have any of the homogeneous solutions in it. How do you identify mathematically the one "true" particular solution?

In practice, of course, it hardly matters. You can pretend there is only one allowable particular solution. But, that assertion doesn't stand up to mathematical scrutiny.
Please tell us just how, where, it fails to stand up to Mathematical scrutiny.
 
  • #10
WWGD said:
Please tell us just how, where, it fails to stand up to Mathematical scrutiny.
Every other particular solution has the same property: that it's a particular solution of the inhomogeneous equation. You can't distinguish them mathematically.
 
  • #12
PeroK said:
That's a false distinction. There's no way to say that ##y_p## doesn't have any of the homogeneous solutions in it. How do you identify mathematically the one "true" particular solution?
I taught classes on differential equations many times. All of the textbooks I've seen distinguish between the particular solution and the general solution, where the general solution consists of the particular solution together with the solution of the homogeneous problem. For a second-order ODE, the general solution will be ##y_g(x) = c_1 f(x) + c_2 g(x)## where the constants are arbitrary if no initial conditions are given.
 
  • #13
Mark44 said:
I taught classes on differential equations many times. All of the textbooks I've seen distinguish between the particular solution and the general solution, where the general solution consists of the particular solution together with the solution of the homogeneous problem. For a second-order ODE, the general solution will be ##y_g(x) = c_1 f(x) + c_2 g(x)## where the constants are arbitrary if no initial conditions are given.
The general solution has two arbitrary coefficients. That's not the argument. The argument is whether there is only one possible choice for the particular solution. Or, whether they are infinitely many choices for a particular solution - and you just pick the first one you find.
 
  • #14
PeroK said:
According to that page, a particular solution is "any solution we can get our hands on".
And Paul distinguishes between a particular solution and the complementary solution.
complementary solution. From the page you cited:
Solving for y(t) gives ##y(t) = c_1y_1(t) + c_2y_2(t) + Y_P(t)##
So clearly he differentiates between the particular solution, general solution, and complementary solution.

@PeroK, can you cite for me any ODE textbook that describes a particular solution as including the complementary solution? It would be foolish to actually do this to find the coefficients of the particular solution function(s), because the complementary portion essentially belongs to the nullspace/kernel of the operator that represents the homogeneous ODE.
 
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  • #15
I agree with @PeroK here. A particular solution is any solution you can get your hands on. Some particular solutions may be easier on the eyes than others, but this

Mark44 said:
The particular solution is what you (@sdfsfasdfasf) showed, namely, ##y_p(x) = \frac 1{12}\sin(3x)##. No other function will satisfy this ODE.
(my emphasis)
is simply false.

Given a particular solution ##y_p(x)## to a linear differential equation, any function on the form
$$
y(x) = y_p(x) + y_h(x)
$$
where ##y_h(x)## is one of the (many) homogeneous solutions will satisfy the original ODE.

Put somewhat more formally: If you have a linear differential operator ##\mathcal L##, want to solve
$$
\mathcal L y = f,
$$
and have access to a function ##y_p## that solves it. Then, for any non-zero solution ##y_h## to the homogeneous equation
$$
\mathcal L y = 0
$$
will give another function ##y = y_p + y_h## to the original differential equation as
$$
\mathcal L (y_p + y_h) = \underbrace{\mathcal L y_p}_{= f} + \underbrace{\mathcal L y_h}_{= 0}
= f + 0 = f.
$$

This is an absolutely fundamental concept in the solving of linear differential equations.
 
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  • #16
@Mark44 In this case, I claim that the function:
$$y_p(x) = \frac{(x + 1)}{12}\sin(3x)$$is also a particular solution of the ODE. Can you explain why this is not a particular solution of the ODE? You might prefer ##\frac 1 {12}x\sin(3x)##, but a personal preference isn't a mathematical property.
 
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  • #17
Mark44 said:
@PeroK, can you cite for me any ODE textbook that describes a particular solution as including the complementary solution?
Okay, you have to stop misquoting me here. I'm really getting fed up with it. Where have I even used the word "complementary"?

Please read what I've actually written!
 
  • #18
Orodruin said:
I agree with @PeroK here. A particular solution is any solution you can get your hands on. Some particular solutions may be easier on the eyes than others, but this


(my emphasis)
is simply false.

Given a particular solution ##y_p(x)## to a linear differential equation, any function on the form
$$
y(x) = y_p(x) + y_h(x)
$$
where ##y_h(x)## is one of the (many) homogeneous solutions will satisfy the original ODE.

Put somewhat more formally: If you have a linear differential operator ##\mathcal L##, want to solve
$$
\mathcal L y = f,
$$
and have access to a function ##y_p## that solves it. Then, for any non-zero solution ##y_h## to the homogeneous equation
$$
\mathcal L y = 0
$$
will give another function ##y = y_p + y_h## to the original differential equation as
$$
\mathcal L (y_p + y_h) = \underbrace{\mathcal L y_p}_{= f} + \underbrace{\mathcal L y_h}_{= 0}
= f + 0 = f.
$$

This is an absolutely fundamental concept in the solving of linear differential equations.
It seems you're saying if x is a solution, then so is x+0. Then 2x=6 when x=3+ k*0 for any k. Isn't this obvious? Does it add anything meaningful to the solution?
 
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  • #19
WWGD said:
It seems you're saying if x is a solution, then so is x+0. Then 2x=6 when x=3+ k*0 for any k. Isn't this obvious? Does it add anything meaningful to the solution?
The point is that you can pick any particular solution. It doesn't matter how you find it or what form it takes. Then you adjust the constants of the homogeneous solution to fit your boundary/initial conditions.

For example, take the ODE
$$
y'' + 4 y = 4.
$$
This has an obvious particular solution ##y_p = 1##, but it also has a particular solution ##y_p = 2\cos^2(x)##. It is fine to use either and write the general solution as
$$
y(x) = y_p(x) + A \cos(2x) + B\sin(2x).
$$
Depending on your choice of particular solution, you will get different values for ##A## if you adapt to the same boundary conditions, but both are perfectly valid particular solutions (even though the first may be easier to work with, but that is besides the point).

Edit: The entire point of finding a single particular solution is that all functions on the form ##y = y_p + y_h## are also solutions to the original ODE.
 
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  • #20
The linear algebra formulation of this is to consider a linear operator ##\mathcal L## on a vector space ##V## and ask which vectors ##x \in V## satisfy ##\mathcal L x = y##. Given any ##x_p## such that ##\mathcal L x_p = y## and any ##x_h \in {\rm ker}\mathcal L##, it holds that ##\mathcal L (x_p + x_h) = y##. There is no a priori way of distinguishing ##x_p## from ##x_p + x_h## in this regard. Both are perfectly valid solutions to ##\mathcal L x = y##.

Then you can start preferring a particular ##x## because it is easier to work with or, if you have an inner product, such that it is the ##x## of the smallest norm, or any other condition you want to apply. It doesn't change the fact that you can pick either as your ##x_p## and write the general solution as ##x_p + x_h## with ##x_h## in the kernel of ##\mathcal L##.
 
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  • #21
Orodruin said:
Edit: The entire point of finding a single particular solution is that all functions on the form ##y = y_p + y_h## are also solutions to the original ODE.
This!!!
 
  • #22
The main lesson here seems to be that when there is an ODE whose solution is
$$y_g = y_h + y_p,$$
##y_g## itself be called a particular solution since, for any ##a\in \mathbb{R}##, ##a \cdot y_h + y_g## is also a solution to the ODE. My question is, would it make sense to define ##y_p## as a solution to the non-homogeneous equation that doesn't contain the term, ##a\cdot y_h##, since it's in the nullspace##^*## of the differential operator that represents the homogeneous ODE (##^*## term introduced in post #14)? I was hoping this would maybe smooth out some of the disagreements between authors and confusion of the OP.

And it was explained that some ODEs have more than one ##y_p##, like shown in post # 16. My question is, does it apply to ODEs with a unique solution, on which you can use an existence and uniqueness theorem, since a unique solution must mean something? As always, thank you for your time and wisdom.
 
  • #23
PeroK said:
In this case, I claim that the function: $$y_p(x) = \frac{(x + 1)}{12}\sin(3x)$$is also a particular solution of the ODE.
Yes. However, by "particular solution," I'm considering only functions that aren't also part of the solution to the homogeneous problem. With that clarification, I still hold that there is only one particular solution that doesn't include a sum of functions, one or more of which is a solution of the homogeneous problem.
PeroK said:
Okay, you have to stop misquoting me here. I'm really getting fed up with it. Where have I even used the word "complementary"?
I'm quoting the term used in the Paul article you posted. "Complementary solution" and "solution of the homogeneous problem" are synonymous.
Orodruin said:
The entire point of finding a single particular solution is that all functions on the form ##y = y_p + y_h## are also solutions to the original ODE.
That's clear and is no different from what I've said all along.

@Orodruin, let's take a look at your example of ##y'' + 4y = 4##, for which a fairly obvious particular solution is ##y_p = 1##. Your alternate particular solution is ##y_p = 2\cos^2(x)##. I agree that this is a particular solution but note that it can also be written as ##y_p = \cos(2x) + 1##. Here the ##\cos(2x)## is a solution of y'' + 4y = 0, while 1 is what I would call the particular solution. Excluding from consideration any function that is a sum of functions in which one or more is a solution of y'' + 4y = 0 from consideration, I still assert that ##y_p = 1## is the only particular solution. So in summary, I would not consider any of the following as particular solutions: ##y_p = 1 + \cos(2x), y_p = 1 + 7\cos(2x), y_p = 1 + 3\cos(2x) + 7\sin(2x),## etc., notwithstanding the fact that all of these are unnecessarily complicated.
 
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  • #24
It seems to me that the main argument here is one of terminology. I still have a couple of ODE textbooks I taught from. One of them, "A First Couse in Differential Equations," by Frank G. Hagin, uses but doesn't define the term "particular solution." The other, "Linear Algebra and Differential Equations," by Charles G. Cullen, defines the general solution of, for example, ##y'' + 4y = 8 + 6\sin(t)## as ##y = c_1 \cos(2t) + c_2\sin(2t) + 2 + 2\sin(t)##, and ##y = -2 \cos(2t) - \frac 3 2 \sin(2t) + 2 + 2\sin(t)## as a particular solution (for a problem with initial conditions).
 
  • #25
Mark44 said:
The other, "Linear Algebra and Differential Equations," by Charles G. Cullen, defines ##y = -2 \cos(2t) - \frac 3 2 \sin(2t) + 2 + 2\sin(t)## as a particular solution (for a problem with initial conditions).
Confusing!
 
  • #26
docnet said:
would it make sense to define ##y_p## as a solution to the non-homogeneous equation that doesn't contain the term, ##a\cdot y_h##, since it's in the nullspace##^*## of the differential operator that represents the homogeneous ODE (##^*## term introduced in post #14)?
No, this is the point of posts #19 and #20. There simply is no unique way to define what ”contain the homogeneous term” means. And some of the ways in which you could do this will not generally give you the ##y_p## you would want (the ”easiest” one) in every single case.

Mark44 said:
However, by "particular solution," I'm considering only functions that aren't also part of the solution to the homogeneous problem.
The point is that this is completely arbitrary and doesn’t really pick out a particular solution.

Mark44 said:
I agree that this is a particular solution but note that it can also be written as ##y_p = \cos(2x) + 1##.
I would like you to note that the function 1 can also be written as ##1 = 2\cos^2(x) - \cos(2x)## where the second term is on the form of the homogeneous solution. Therefore, your own argument excludes using 1 as the particular solution as it can be written as a particular solution plus a homogeneous solution. This is the point. You simply cannot pick out a single particular solution just because you like the form it is written on better.

Mark44 said:
Excluding from consideration any function that is a sum of functions in which one or more is a solution of y'' + 4y = 0 from consideration
So ##1 = 2\cos^2(x) - \cos(2x)## is not a particular solution as it has a term which is a solution to the homogeneous equation. Got it!
 
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  • #27
Orodruin said:
The linear algebra formulation of this is to consider a linear operator ##\mathcal L## on a vector space ##V## and ask which vectors ##x \in V## satisfy ##\mathcal L x = y##. Given any ##x_p## such that ##\mathcal L x_p = y## and any ##x_h \in {\rm ker}\mathcal L##, it holds that ##\mathcal L (x_p + x_h) = y##. There is no a priori way of distinguishing ##x_p## from ##x_p + x_h## in this regard. Both are perfectly valid solutions to ##\mathcal L x = y##.

Then you can start preferring a particular ##x## because it is easier to work with or, if you have an inner product, such that it is the ##x## of the smallest norm, or any other condition you want to apply. It doesn't change the fact that you can pick either as your ##x_p## and write the general solution as ##x_p + x_h## with ##x_h## in the kernel of ##\mathcal L##.
A genuine question if I may ask, couldn't one always distinguish ##x_p## from ##x_p+x_h## if you knew what ##x_h## was?
 
  • #28
docnet said:
A genuine question if I may ask, couldn't one always distinguish ##x_p## from ##x_p+x_h## if you knew what ##x_h## was?
No. See above. Any two functions that solve the differential equation will have a difference that is a homogeneous solution. Which you choose to write as the other plus a homogeneous part is completely arbitrary.
 
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  • #29
Orodruin said:
I would like you to note that the function 1 can also be written as ##1 = 2\cos^2(x) - \cos(2x)## where the second term is on the form of the homogeneous solution. Therefore, your own argument excludes using 1 as the particular solution as it can be written as a particular solution plus a homogeneous solution. This is the point. You simply cannot pick out a single particular solution just because you like the form it is written on better.


So ##1 = 2\cos^2(x) - \cos(2x)## is not a particular solution as it has a term which is a solution to the homogeneous equation. Got it!
Really, why would anyone consider doing this? My approach to this kind of problem is a set of basis functions for the homogeneous problem, and then find a function that satisfies the nonhomogeneous problem. So for the problem at hand, the complementary solution is some linear combination of ##\sin(2x)## and ##\cos(2x)##. Inasmuch as ##y_p = 1## satisfies y'' + 4y = 4, I would definitely call that a particular solution. Further, there is no linear combination of ##\sin(2x)## and ##\cos(2x)## such that ##\mathcal L(c_1\sin(2x) + c_2\cos(2x)) = 4##, to borrow your operator notation.
Orodruin said:
The linear algebra formulation of this is to consider a linear operator
##\mathcal L## on a vector space ##V## and ask which vectors ##x \in V## satisfy ##\mathcal L x = y##. Given any ##x_p## such that ##\mathcal L x_p = y## and any ##x_h \in {\rm ker}\mathcal L##, it holds that ##\mathcal L (x_p + x_h) = y##.
Clearly.
Orodruin said:
There is no a priori way of distinguishing ##x_p## from ##x_p + x_h## in this regard. Both are perfectly valid solutions to ##\mathcal L x = y##.
Right, I can't distinguish ##x_p## from ##x_p + x_h## but I can distinguish ##x_p## from ##x_h## by virtue of the fact that ##\mathcal L(x_h) = 0## and ##\mathcal L(x_p) = y##. ##x_h \in {\rm ker}\mathcal L## while ##x_p \notin {\rm ker}\mathcal L##.
Orodruin said:
Then you can start preferring a particular ##x## because it is easier to work with
This -- taking a cue from Occam's Razor.
 
  • #30
Mark44 said:
Really, why would anyone consider doing this?
That is really not the question you should be asking yourself. The point is that they are both particular solutions and it doesn’t matter for the solution which one you use. Both are particular solutions. Being a particular solution only means satisfying the differential equation without having undetermined constants. That’s it. The example in question is a bit pathological as one of the particular solutions is 1, but it can easily happen that both particular solutions are on similar levels of aesthethics.

Then it is a question from problem to problem which particular solution is actually the easier one to find and use - but any particular solution will do.

Mark44 said:
Inasmuch as yp=1 satisfies y'' + 4y = 4, I would definitely call that a particular solution.
So does ##2\cos^2(x)##. Meaning it is a particular solution as it does not contain undetermined constants.

Mark44 said:
This -- taking a cue from Occam's Razor.
But being easier to work with does not mean that other solutions are not particular solutions. Just that they are easier to work with.


Mark44 said:
Right, I can't distinguish ##x_p## from ##x_p + x_h##

But this is what you are trying to do by claiming that there only exists one particular solution!


Mark44 said:
but I can distinguish ##x_p## from ##x_h## by virtue of the fact that ##\mathcal L(x_h) = 0## and ##\mathcal L(x_p) = y##. ##x_h \in {\rm ker}\mathcal L## while ##x_p \notin {\rm ker}\mathcal L##.
Which is irrelevant here as it is not what is being discussed.
 
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  • #31
Well this post is doing well!
I need some further clarification. For example, in my textbook I am given a 2nd order inhomogeneous differential equation, from which I am able to solve the auxiliary equation, and find a, not the particular solution, and clearly the particular solution I choose cannot have a part of the inhomogeneous solution in it (could someone explain why please!). Now, I am also given some initial conditions, my question to all of you is, what do I call the final answer? It contains no undetermined constants, clearly it solves the differential equation, and it contains a part of the homogeneous solution. Is particular solution a good name for it? I struggle to justify that name. Also, is there only one function f which solves the differential equation and has the required initial condititions? All throughout my textbook it states things like given these (conditions) find the particular solution, that seems to go against everything I have read here.
 
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  • #32
Mark44 said:
Yes. However, by "particular solution," I'm considering only functions that aren't also part of the solution to the homogeneous problem.
That has no meaning in general. In this case, I just explicitly added a term in ##\sin 3x##, but for a different ODE, it would less obvious where and when you can subtract a solution to the homogeneous equation. In any case, I claim that the particular solution is:

Any function which is a solution to the inhomogeneous ODE.

Whereas, it's not clear what your mathematical definition is. How do you intend to uniquely identify this function from its mathematical properties? You need to specify what is unique about it.

The particular solution to a inhomogeneous ODE is the unique solution, which ...?

Also, the choice of ##\cos(3x)## and ##\sin(3x)## as the basis for your solution space is arbitrary. That basis emerges from a standard technique, but it doesn't make those two functions the only functions with that property. In fact, if we extend things to allow complex solutions, then we have a common alternative in terms of complex exponentials: ##e^{3ix}## and ##e^{-3ix}##. Even without complex numbers, there are infinitely many pairs of functions that span the null space.

This is an important point for the the student, as there are many cases in theory and application where the arbitrariness of the functions you choose to represent the null space and the particular solution is an important concept.

Also, just to repeat the quotation from Paul's online notes:

The particular solution is "any solution we can get our hands on". That is very different from your belief that there is a unique choice of particular solution.
 
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  • #33
sdfsfasdfasf said:
Well this post is doing well!
Sorry for the unseemly squabbling.
sdfsfasdfasf said:
I need some further clarification. For example, in my textbook I am given a 2nd order inhomogeneous differential equation, from which I am able to solve the auxiliary equation, and find a, not the particular solution, and clearly the particular solution I choose cannot have a part of the inhomogeneous solution in it (could someone explain why please!).
I don't understand what you mean here.
sdfsfasdfasf said:
Now, I am also given some initial conditions, my question to all of you is, what do I call the final answer? It contains no undetermined constants, clearly it solves the differential equation, and it contains a part of the homogeneous solution. Is particular solution a good name for it?
As I said above, in Paul's online notes he calls this the "actual" solution. I'm not sure whether looking for standard terminology at this level is a good idea. Authors are going to use different terminology and you are going to have to deal with it.

I might say it is the solution to the IVP (initial value problem).
sdfsfasdfasf said:
I struggle to justify that name. Also, is there only one function f which solves the differential equation and has the required initial condititions? All throughout my textbook it states things like given these (conditions) find the particular solution, that seems to go against everything I have read here.
I think that is using "particular" in a different semantic context and should be avoided. As it's already standard terminology for a chosen solution to the general inhomogeneous equation.

Textbook authors are only human and you will find that sort of minor inconsistency here and there.
 
  • #34
PeroK said:
Also, the choice of ##\cos(3x)## and ##\sin(3x)## as the basis for your solution space is arbitrary. That basis emerges from a standard technique, but it doesn't make those two functions the only functions with that property. In fact, if we extend things to allow complex solutions, then we have a common alternative in terms of complex exponentials: ##e^{3ix}## and ##e^{-3ix}##. Even without complex numbers, there are infinitely many pairs of functions that span the null space.
Here is another example, where the choice of particular solution does not have a clear ”simplest” form:

Consider ##y’ + y = e^x##. Since ##de^x/dx = e^x##, this has a particular solution ##y_p = e^x/2##. However, other particular solutions include ##y_p = \sinh(x)## and ##y_p = \cosh(x)##, which are just as viable.
 
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  • #35
With regard to ##y'' + 9y = \frac 1 2 \cos(3x)##:
PeroK said:
I claim that the function $$y_p(x) = \frac{(x + 1)}{12}\sin(3x)$$is also a particular solution of the ODE. Can you explain why this is not a particular solution of the ODE?
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?
PeroK said:
Also, the choice of ##\cos(3x)## and ##\sin(3x)## as the basis for your solution space is arbitrary. That basis emerges from a standard technique, but it doesn't make those two functions the only functions with that property. In fact, if we extend things to allow complex solutions, then we have a common alternative in terms of complex exponentials: ##e^{3ix}## and ##e^{-3ix}##. Even without complex numbers, there are infinitely many pairs of functions that span the null space.
Of course, a basis for the solution space isn't unique, and I don't think I ever said anything to the contrary.
Mark44 said:
Right, I can't distinguish ##x_p## from ##x_p + x_h##
Orodruin said:
But this is what you are trying to do by claiming that there only exists one particular solution!
So far, the examples of particular solutions that violate my claim appear to be 1) those that are a sum of a viable solution to the nonhomogeneous problem plus a function that is some linear combination of basis functions that are solutions to the homogeneous problem. E.g. ##y_p = \frac x {12}\sin(3x) + \frac 1 {12}\sin(3x)##, or 2) functions that are identically equal to some solution of the nonhomogeneous problem. E.g. ##y_p = 1## vs. ##y_p = 2\cos^2(x) - \cos(2x)## (relative to y'' + 4y = 4).
Mark44 said:
but I can distinguish ##x_p## from ##x_h## by virtue of the fact that ##\mathcal L(x_h) = 0## and ##\mathcal L(x_p) = y##. ##x_h \in {\rm ker}\mathcal L## while ##x_p \notin {\rm ker}\mathcal L##.
Orodruin said:
Which is irrelevant here as it is not what is being discussed.
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.
Orodruin said:
Here is another example, where the choice of particular solution does not have a clear ”simplest” form:
Consider ##y’ + y = e^x##. Since ##de^x/dx = e^x##, this has a particular solution ##y_p = e^x/2##. However, other particular solutions include ##y_p = \sinh(x)## and ##y_p = \cosh(x)##, which are just as viable.
With trig or hyperbolic trig functions, one can find other functions that are identically equal. The example you gave of ##1 = 2\cos^2(x) - \cos(2x)## falls into this category. Can you argue that the two sides of this equation are actually different functions?
 

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