# General solutions of linear differential equations, why add the homogen. + particular

by DrummingAtom
Tags: differential, equations, homogen, linear, solutions
 P: 661 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
 Engineering Sci Advisor HW Helper Thanks P: 7,280 "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.
 P: 661 Yeah, I'm just looking for something deeper but I guess it stops there. Thanks for your reply.
P: 3,313
General solutions of linear differential equations, why add the homogen. + particular

 Quote by DrummingAtom 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.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 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 [b]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.
 P: 661 @ 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. Attached Thumbnails
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