Determining linear independence of three simple functions for a third order ODE. (2.9-6)

Question

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.

1. Did I actually answer/solve all parts of this problem?
2. Determinants of derivatives can get messy. Is there a less tedious more efficient approach to this?
3. I need a little help with how to select an open interval $I$ by which a basis is established.
4. I used the matrix TEX commands to form my determinant. Does a TEX for determinants exist?
5. 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?

Answer 1

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.

Answer 2

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.