Why Add Homogeneous and Particular Solutions in Linear Differential Equations?

[DrummingAtom]

I've come across an issue that was bugging me last semester in my circuits class today: Finding general solutions of linear differential equations. Just add the homogeneous and particular solution and it's done. Last semester it wasn't explained why exactly this is possible and this semester it's even worse.

Is there a proof or preferably a geometric representation that shows why adding these two together works? I can't find anything in my Diffy Q book or online. The only thing that is said is along the lines of "because they are linear." I guess I'm just not finding that point very obvious.. Thanks for any help

 

[AlephZero]

"Because they are linear" means that if y = f(x) and y = g(x) are solutions of the equation, then any linear combination y = a f(x) + b g(x) is also a solution.

So, if can you find one particular solution of the complete equation (without any arbitrary constants), usually with a certain amount of "guessing" what the terms will be in the solution, and then find the general solution of the homogeneous equation (with the right hand side deleted and set to 0), any linear combination of the two is also a solution of the complete equation, and it also contains the right number of arbitrary constants to be "the" general solution.

 

[DrummingAtom]

Yeah, I'm just looking for something deeper but I guess it stops there. Thanks for your reply.

 

[Stephen Tashi]

Yeah, I'm just looking for something deeper.

Actually, it would be interesting to go deeper. The place where the water gets deeper is the phrase

it also contains the right number of arbitrary constants to be "the" general solution.

It's easy (I think) to accept that a general solution plus a particular solution is a solution. It's harder to see why it expresses all solutions. If you accept that all solutions are found by a certain integration then you can probably accept that only one constant of integration is involved. But since the practical method of solving a differential equation isn't always by a straightforward integration, it's hard to see why we can rely on integration to count the number of possible constants.

 

[HallsofIvy]

If you want a "deeper" answer, look to Linear Algebra and the distinction between a "subspace" and a "linear manifold" of a vector space. A subspace of a vector space is a subset, A, such that if u and v are in A then so are u+ v and av for any scalar a. If, instead, A is a "linear manifold", that is not true but there must exist a subspace A' and fixed vector, r, so that for any vector v in A, v- r is in A'. Conversely, if p is a vector in linear manifold A, then p- r is in subspace A'. In particular, if p and q are vectors in A, then there difference, p- q, is in the subspace A'.

In R², any subspace is a straight line through the origin. A "linear manifold" is a straight line that is not through the origin. The vector, r, above, can be any vector from the origin to a point on the line. Any point on that line can be written as r plus a point on the line parallel to that given line, thorugh the origin- which is, of course, a subspace.

The fundamental "theory" of linear differential equations is that the set of all solutions to a linear homogeneous differential equation form a subspace of the space of all infinitely differentiable functions. The set of solutions to the corresponding homogeneous equation is the "line through the origin" and the "particular soution" is the vector "r" from the origin to the linear manifold.

 

[DrummingAtom]

@ Stephen Tashi: That's actually something that I never thought about it. Interesting

@ HallsofIvy: Yes, that's exactly what I'm looking for. But I still don't understand how the homogeneous solution plus "r" would still form a subspace. It seems that adding the particular solution pops the solutions out of the subspace. I attached some pictures to show you how I'm thinking about this. If I drew that correctly then the general solution can't be multiplied by a negative scalar because it the line starts on the origin instead of going through it.

 

[HallsofIvy]

@ Stephen Tashi: That's actually something that I never thought about it. Interesting

@ HallsofIvy: Yes, that's exactly what I'm looking for. But I still don't understand how the homogeneous solution plus "r" would still form a subspace.k

Well, the problem I have with answering this is that I just told you it doesn't form a subspace!

It seems that adding the particular solution pops the solutions out of the subspace. I attached some pictures to show you how I'm thinking about this. If I drew that correctly then the general solution can't be multiplied by a negative scalar because it the line starts on the origin instead of going through it.

Your picture looks good. Yes, it is true that the general solution can't be multiplied by a scalar whether positive or negative (and give another solution) because the general solution does NOT form a subspace.

 

[DrummingAtom]

Oh, my mistake, that didn't make much sense because you explicitly said what is going on, lol. Subtracting "r" from the general solution would form a subspace, or that A' you were talking about earlier, which is the set of the homogeneous solutions. Thanks for your insight.

 

[LCKurtz]

I will throw my 2 cents worth. Let's call your DE $L(y) = f(x)$. In the second order case you have two linearly independent solutions $y_1$ and $y_2$, giving the general solution $y_c= C_1y_1 + C_2y_2$ of the homogeneous equation $L(y) = 0$. Now let's say you have found, by any method, some particular solution $y_p$ of the NH equation so $L(y_p) = f(x)$. The claim is then that the general solution of the NH equation is $y_g = C_1y_1+C_2y_2 + y_p$. To see this, suppose $Y$ is any solution of the NH equation. Then $L(Y) = f(x)$. Look at what $Y-y_p$ satisfies. $L(Y-y_p) = L(Y) - L(y_p) = f(x) - f(x) = 0$ by the linearity of $L$. This says $Y-y_p$ is a solution of the homogeneous equation. Therefore it must be possible to express $Y - y_p = C_1y_1 + C_2y_2$ so $Y = C_1y_1+C_2y_2 + y_p$. So just having $y_p$ is good enough that you can get any solution, such as $Y$, by adding it to the general solution of the homegeneous DE.

I know this information is contained in the previous replies, but I hope you find this helpful.