# Basis of the solution space of a differential equation

1. Apr 7, 2015

### Kavorka

1. The problem statement, all variables and given/known data
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.

3. 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.

2. Apr 7, 2015

### Staff: Mentor

Yes.
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.

3. Apr 7, 2015

### 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)

4. Apr 8, 2015

### pasmith

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 $c_1 = c_2 = 0$.

5. Apr 8, 2015

### Staff: Mentor

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 $c_2 = 0$.

6. Apr 8, 2015

### Kavorka

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

7. Apr 8, 2015

### Staff: Mentor

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

8. Apr 8, 2015

### 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.

9. Apr 8, 2015

### Staff: Mentor

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.

Last edited: Apr 9, 2015