General solution vs particular solution

[sdfsfasdfasf]

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.

 

[PeroK]

There's no such thing as "the" particular solution. There is only "a" particular solution. Any one will do.

 

[sdfsfasdfasf]

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.

 

[Mark44]

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 $c_2$.

 

[PeroK]

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.

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.

 

[PeroK]

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.

 

[Mark44]

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.

 

[PeroK]

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.

 

[WWGD]

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.

 

[PeroK]

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.

 

[PeroK]

As always, Paul has the answer:

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

According to that page, a particular solution is "any solution we can get our hands on".

 

[Mark44]

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.

 

[PeroK]

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.

 

[Mark44]

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.

 

[Orodruin]

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

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.

 

[PeroK]

@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.

 

[PeroK]

@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!

 

[WWGD]

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?

 

[Orodruin]

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.

 

[Orodruin]

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 $\mathcal L$.

 

[PeroK]

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!!!

 

[docnet]

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.

 

[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.

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.

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.

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.

 

[Mark44]

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).

 

[docnet]

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!

 

[Orodruin]

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.

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.

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.

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!

 

[docnet]

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?

 

[Orodruin]

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.

 

[Mark44]

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.

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.

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 $x_p \notin {\rm ker}\mathcal L$.

Then you can start preferring a particular $x$ because it is easier to work with

This -- taking a cue from Occam's Razor.

 

[Orodruin]

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.

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.

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.

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!

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.