# Determining # of Solutions for DEs

1. Oct 22, 2014

### MathewsMD

For a certain DE, how can we be certain how many solutions there are? Also, how can we determine the general set?

For example, for the linear, second order, homogeneous DE, you may have ay'' + by' + cy = 0 [1] and then the characteristic equation of the form ar^2 + br + c can be found, and thus the corresponding r values to yield the general solution of the form: c1y1 + c2y2 where y1 = e^(r1t) and y2 = e^(r2t). For this, you find the Wronskian, and if non-zero, then you know these solutions are independent. That's great. And if anything up to here is incorrect, please correct me.

But my question is: how are we CERTAIN c1y1 + c2y2 contains every possible solution to [1]? I understand this is a solution, but don't see why this encompasses every single solution. Likewise (I assume it is corollary), why does an nth order, linear, homogeneous DE have n terms in the general solution?

2. Oct 22, 2014

### SteamKing

Staff Emeritus
The answer to this question gets into the theory of ordinary differential equations, especially dealing with the existence and uniqueness of solutions.

The number of solutions to an ODE is related to the order of the equation, since ODEs having an order greater than 1 can be reduced to a system of n first-order equations, where n is the order of the original equation. Since a first order equation has a single solution to within an additive constant, an n-th order equation can have at most n different solutions.

http://en.wikipedia.org/wiki/Ordinary_differential_equation

At least, that's my understanding of the theory of ODEs.

3. Oct 22, 2014

### Staff: Mentor

To add to what SteamKing said, the order of a differential equation is equal to the dimension of the function space of solutions. First order ODE: dimension is 1; second order ODE: dimension is 2, and so on for third- and higher-order ODEs.

For second order ODEs, if you have two solutions that are linearly independent, they constitute a basis for the solution space; hence every solution is a linear combination of the two functions in the basis. IOW, if y1 and y2 are linearly independent solutions, then every solution of this ODE can be written as c1y1 + c2y2, for some constants c1 and c2.

The function space concept is nearly identical to the linear algebra concept of a vector space. In $\mathbb{R}^2$, a two-dimensional vector space, if you have two linearly independent vectors, then those vectors form a basis for this space, and every vector in this space is some linear combination of the two basis vectors. It's pretty much the same for a two-dimensional function space, such as the space of solutions of a second order ODE.

Last edited: Oct 22, 2014
4. Oct 29, 2014

### HallsofIvy

The phrase "the dimension of the function space of solutions" applies only to linear differential equations. Sets of solutions to non-linear differential equations are not, in general, linear spaces and so have no "dimension".

5. Oct 29, 2014

### MathewsMD

Oh really? Do you mind expanding? It's all very interesting but if you could provide a bit of intuition behind why, that would be greatly appreciated!

6. Oct 30, 2014

### HallsofIvy

Expand what? Do you know what "linear vector spaces" (Linear Algebra) are? Do know how to show that the set of all solutions to a "linear homogeneous differential equation" forms a vector space? If so you should be able to see why sets of solutions to non-linear differential equations do NOT form vector spaces.