Is our complementary solution guaranteed to be linearly independent?

[dumbQuestion]

I had kind of a general question. Say I have a second order, homogeneous ODE. Say I use one of the general techniques to generate a complementary solution for my ODE and I end up with something of the form y = C1(solution1) + C2(solution2)

Am I gauranteed that these two solutions will be linearly independent or do I need to verify with the Wronskian each time?


Also, can I think of linearly independent solutions as a "basis" like I do in linear algebra? It seems like it because its my understanding that any solution for the ODE is simply a linear combination of solutions in this "basis"


Also, can we tell the dimensions of this basis based on the order of the ODE? Like, a second order ODE has two solutions in its basis, a third order has 3, a first order has one, etc.


Thanks!

 

[Muphrid]

Yes, you can think of the solutions as a basis in a vector space. In this case, they're basis functions in a space of functions, but most of the ideas from vector algebra still apply.

I think often the techniques chosen to generate solutions are meant to come up with solutions out of orthogonal functions--sines and cosines, for instance, are automatically linlearly independent as long as none of their frequencies are the same. Someone may know more on this matter, though.

 

[HallsofIvy]

A complementary solution can't be "dependent" on the solution to the homogeneous equation. Suppose the equation is L(f)= g(x) where L is the linear differential operator, u(x) a linear combination of the independent solutions to the homogenelus equation, and v(x) a complementary solution. To be dependent means that there exist numbers, a and b, non-zero, such that au(x)+ bv(x)= 0. Applying L to both sides of that, L(au(x)+ bv(x))= aL(u)+ bL(v)= a(0)+ b(g)= 0 so that bg(x)= 0 for all x. Since g is not always 0, b= 0. That makes our original equation au(x)= 0 for all x and so a= 0.

The solutions to any linear homogeneus nth order differential equation form a vector space of dimension n. The solutions to a non-homogeneous linear equation do NOT form a vector space but do form a "linear manifold". If you think of a one-dimensional vector space as a line through the origin of a xy-coordinate system, or a two dimensional vector space as a plane through the origin of a xyz-coordinate system, then you can think of a one dimensional "linear manifold" as a line NOT through the origin or a two dimensional "linear manifold" NOT through the origin.

 

[dumbQuestion]

If you think of a one-dimensional vector space as a line through the origin of a xy-coordinate system, or a two dimensional vector space as a plane through the origin of a xyz-coordinate system, then you can think of a one dimensional "linear manifold" as a line NOT through the origin or a two dimensional "linear manifold" NOT through the origin.

Thank you, this really does help me visualize this.So I want to make sure I understand. So for example, when dealing with non homogeneous equations, does it mean we can't establish a "basis" or fundamental set of solutions so that every particular solution can be expressed as a linear combination of those solutions?THe thing that prompted me to ask was this particular problem:y'' - 5y' + 4y = 8e^x

I can generate a basis for the vector space of solutions for the associated homogeneous equation and its B = {e^4x,e^x} (I got this by solving the auxilary equation and attaining distinct real roots x=4 and x=1)

Using variation of parameters, I can then generate a particular solution for this sytem:

y_p=(-8/9)e^x-(8/3)xe^xTo make sure it is indeed a particular solution I took y'_p and y''_p and plugged it into the original ODE and it does check out. However, y_p is not a linear combination of the functions in B.

 

[dumbQuestion]

oh wait I think I get it, the "basis" as I'm thinking of it for a nohomogeneous linear 2nd order ODE would be B={b₁,b₂,y_p}where b₁ and b₂ are the functions in the basis for the associated homogeneous ODE and y_p is a particular solution satisfying the non homogeneous ODE.EDIT: Oh waaaait... No what I just wrote is wrong. Any solution Y(x) of the non homogeneous system can be expressed as:

Y=C₁b₁+C₂b₂+y_pSo I can see why {b₁,b₂,y_p} is NOT a "basis" because we can't just have any linear combination of these to generate solutions, the solutions are linear combinations of b₁ and b₂ with y_p simply added on (not any multiple of y_p)! So this is why it's not a vector space but a linear manifold as you described it