Basis of the solution space of a differential equation

[Kavorka]

Homework Statement

Verify that the functions y₁(x) = x and y₂(x) = 1/x are solutions of the differential equation y'' + (1/x)y' - (1/x²)y = 0 on I = (0,∞).

Show that y₁(x), y₂(x) is a basis of the solution space of the differential equation.

The Attempt at a Solution

For the first part I'll just plug the functions back into the differential equation and state the interval in terms of domains, not hard. I'm just confused about how to show the second part.
Do I need to show that the two solutions are linearly independent and span the solution space? How do I do this? It's probably simpler than I would think it's just that the language is confusing, mixing linear algebra and differential equations. The general solution would be stated as y(x) = c₁y₁(x) + c₂y₂(x). Does this in itself show that the functions form a basis? Since a linear combination of the functions resulted in the solution space? Confusing.

 

[Mark44]

Homework Statement

Verify that the functions y₁(x) = x and y₂(x) = 1/x are solutions of the differential equation y'' + (1/x)y' - (1/x²)y = 0 on I = (0,∞).

Show that y₁(x), y₂(x) is a basis of the solution space of the differential equation.

The Attempt at a Solution

For the first part I'll just plug the functions back into the differential equation and state the interval in terms of domains, not hard. I'm just confused about how to show the second part.
Do I need to show that the two solutions are linearly independent and span the solution space?

Yes.

How do I do this? It's probably simpler than I would think it's just that the language is confusing, mixing linear algebra and differential equations.

The dimension of the solution space is 2, since your DE is second order. Just show that the two solutions are linearly independent using the familiar definition of linear independence; i.e., that the equation $c_1y_1 + c_2y_2 = 0$ has only one solution in terms of the constants.

The general solution would be stated as y(x) = c₁y₁(x) + c₂y₂(x). Does this in itself show that the functions form a basis? Since a linear combination of the functions resulted in the solution space? Confusing.

 

[fourier jr]

for linear independence you could also find out whether the wronskian is 0 or not: $det\left( \begin{matrix} f_{1} & f_{2} \\ f_{1}^{\prime} & f_{2}^{\prime} \end{matrix}\right)$

(0 means the solutions are linearly dependent)

 

[pasmith]

Just show that the two solutions are linearly independent using the familiar definition of linear independence; i.e., that the equation $c_1y_1 + c_2y_2 = 0$ has only one solution in terms of the constants.

It is not sufficient that there be only one solution for c_1 and c_2; it is also necessary that the only solution be c_1 = c_2 = 0.

 

[Mark44]

It is not sufficient that there be only one solution for c_1 and c_2; it is also necessary that the only solution be c_1 = c_2 = 0.

If there is only one solution in terms of the constants, it follows that the solution is necessarily the trivial solution; i.e., $c_1 = 0$ and $c_2 = 0$.

 

[Kavorka]

Is showing linear independence of the functions all I have to do to show that they form a basis?

 

[Mark44]

Is showing linear independence of the functions all I have to do to show that they form a basis?

You should also mention that the dimension of the solution space is 2, so any basis for it can consist of only two vectors

 

[Kavorka]

Let me just make sure I understand this. The dimension of the solution space of a differential equation will be the highest derivative of the equation, in this case 2. If the functions are linearly independent within the solution space (in this case (0,∞)), because there are 2 of them in a 2-D solution space and they are linearly independent, they span the solution space. Thus, they form a basis.

 

[Mark44]

Let me just make sure I understand this. The dimension of the solution space of a differential equation will be the highest derivative of the equation, in this case 2. If the functions are linearly independent within the solution space (in this case (0,∞)), because there are 2 of them in a 2-D solution space and they are linearly independent, they span the solution space. Thus, they form a basis.

The solution space is not (0, ∞), which is a one-dimensional subset of R. The solution space is the set of all linear combinations of the basis functions. If f and g are two linearly independent solutions of your diff. equation, the solution space is ${c_1f + c_2g}$.

What I'm calling the solution space is also called a function space, the set of all linear combinations of some basis set, which in this case is a set of functions. A function space is one kind of vector space, but instead of vectors, we have functions. The axioms of a vector space also apply to a function space.