Determining # of Solutions for DEs

  • Thread starter MathewsMD
  • Start date
433
7
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?
 

SteamKing

Staff Emeritus
Science Advisor
Homework Helper
12,794
1,665
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?
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.
 
32,631
4,377
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:

HallsofIvy

Science Advisor
Homework Helper
41,698
871
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".
 
433
7
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".
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!
 

HallsofIvy

Science Advisor
Homework Helper
41,698
871
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.
 

Related Threads for: Determining # of Solutions for DEs

  • Posted
Replies
3
Views
5K
  • Posted
Replies
8
Views
2K
Replies
4
Views
1K
  • Posted
Replies
3
Views
2K
Replies
2
Views
2K
Replies
8
Views
2K
  • Posted
Replies
3
Views
1K
  • Posted
Replies
0
Views
2K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top