Determining linear independence of three simple functions for a third order ODE. (2.9-6) I would like a sharp independent eye other than my own to review my work here. I have a few questions I would like answered.


*

*Did I actually answer/solve all parts of this problem?

*Determinants of derivatives can get messy. Is there a less tedious more efficient approach to this?

*I need a little help with how to select an open interval $I$ by which a basis is established.

*I used the matrix TEX commands to form my determinant. Does a TEX for determinants exist?

*How do I create a dot (instead of asterisk) for multiplication?


Problem in text:
I need to show that the given functions form a basis of solutions of the given differential equation on any open interval $I$. But first I must use the Wronskian determinant to verify linear independence.
$$y_1=e^{-x}; y_2=xe^{-x}; y_3=x^2e^{-x}$$
$$y'''+3y''+3y'+y=0$$
My Solution:
$$W =\begin{bmatrix} e^{-x} & xe^{-x} & x^2e^{-x}\\-e^{-x} & e^{-x}-xe^{-x} & 2xe^{-x}-x^2e^{-x}\\e^{-x} & -2e^{-x}+e^{-x} & 2e^{-x}-2xe^{-x}-2xe^{-x}+x^2e^{-x}\\\end{bmatrix}$$
$$W =\begin{bmatrix} e^{-x} & xe^{-x} & x^2e^{-x}\\-e^{-x} & e^{-x}-xe^{-x} & 2xe^{-x}-x^2e^{-x}\\e^{-x} & -2e^{-x}+xe^{-x} & 2e^{-x}-4xe^{-x}+x^2e^{-x}\\\end{bmatrix}$$
Below row 2 added to row 1 and row 2 added to row 3:
$$W =\begin{bmatrix} 0 & e^{-x} &  2xe^{-x}\\-e^{-x} & e^{-x}-xe^{-x} & 2xe^{-x}-x^2e^{-x}\\0 & -e^{-x} & 2e^{-x}-2xe^{-x}\\\end{bmatrix}$$
Below col 1 added to col 2:
$$W =\begin{bmatrix} 0 & e^{-x} &  2xe^{-x}\\-e^{-x} & -xe^{-x} & 2xe^{-x}-x^2e^{-x}\\0 & -e^{-x} & 2e^{-x}-2xe^{-x}\\\end{bmatrix}$$
Below row 3 added to row 1:
$$W =\begin{bmatrix} 0 & 0 &  2e^{-x}\\-e^{-x} & -xe^{-x} & 2xe^{-x}-x^2e^{-x}\\0 & -e^{-x} & 2e^{-x}-2xe^{-x}\\\end{bmatrix} = e^{-x}\begin{bmatrix} & 0 &  2e^{-x} \\ & -e^{-x} & 2e^{-x}-2xe^{-x}\\\end{bmatrix}$$
$$W=e^{-x}(-1)(-e^{-x})(2e^{-x})=2e^{-3x}$$
Linear independence has been verified for all $x$ in $I$. I think?
Now let us superposition all functions into one and take derivatives:
$$y= e^{-x} + xe^{-x} + x^2e^{-x}$$
$$y'= xe^{-x} - x^2e^{-x}$$
$$y''= e^{-x} - 3xe^{-x} + x^2e^{-x}$$
$$y'''= -4e^{-x} + 5xe^{-x} - x^2e^{-x}$$
We are now ready to substitute functions to establish basis in our given differential equation:
$$y'''+3y''+3y'+y=0$$
$$(-4e^{-x} + 5xe^{-x} - x^2e^{-x})+3(e^{-x} - 3xe^{-x} + x^2e^{-x})+3(xe^{-x} - x^2e^{-x})+(e^{-x} + xe^{-x} + x^2e^{-x})=0$$
$$0*e^{-x} + 0*xe^{-x} + 0*x^2e^{-x}=0$$
This appears to be the solution. I think?
 A: There's a problem in the second part of your proof, where you try to show it's a basis of the solution space. In general, there are two strategies you can take: prove that each individual solution is actually a solution to the equation, or prove that an arbitrary linear combination of the solutions is itself a solution. You proved that a specific linear combination (namely $y_1+y_2+y_3$) is a solution. That's not enough.
For what it's worth, I recommend testing each $y_i$ individually. It's usually easier than dealing with an arbitrary linear combination.
In answer to your second question, there's a much easier way to prove linear independence that doesn't rely on the Wronskian.
Notice $c_1y_1+c_2y_2+c_3y_3=e^{-x}(c_1+c_2x+c_3x^2)$. We know $1$, $x$, and $x^2$ are linearly independent (they are the standard basis for the vector space of quadratic polynomials), and $e^{-x}$ is never zero, so the solutions are linearly independent.
I also would recommend memorizing the form of the determinant for a 3x3 matrix, so you don't have to mess about with row operations. There's a neat trick to it that involves summing the product of the diagonals with alternating signs. I suggest you look it up; it will save you time in the long run.
A: you could make your life easy if you recognize that $\{e^{-x}, xe^{-x}, x^2e^{-x} \}$ is linearly independent if and only if $\{1, x, x^2 \}$ now do the wronskian if you must or use the fact if a polynomial is identically zero, then it must be the zero polynomial.
