The vectors in
form a basis of
if they are mutually linearly independent and span
.
To check for independence, we can compute the Wronskian determinant:

The determinant is non-zero, so the vectors are indeed independent.
To check if they span
, you need to show that any vector in
can be expressed as a linear combination of the vectors in
. We can write an arbitrary vector in
as

Then we need to show that there is always some choice of scalars
such that

This is equivalent to solving

or the system (in matrix form)

This has a solution if the coefficient matrix on the left is invertible. It is, because

(that is, the coefficient matrix is not singular, so an inverse exists)
Compute the inverse any way you like; you should get

Then


A solution exists for any choice of
, so the vectors in
indeed span
.
The vectors in
are independent and span
, so
forms a basis of
.