This is a very similar post to one previous by me but I felt that not all questions were satisfactorily answered. But I am sincerely grateful to those who tried.
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 higher order (greater than 2) derivatives can be messy. Is there a less tedious more efficient approach to this? Perhaps I should have factored out the common factor $e^x$ before solving it?
- Please pay special attention to how I established basis for my given interval $I$. I believe this may be my weakness.
- I used the matrix TEX tag to form my determinants. Does an actual TEX tag for determinants exist?
- How do I create a TEX dot (instead of asterisk) for multiplication or dot product?
Problem in text:
I need to show that the given functions form a basis of solutions for 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:
$$W =\begin{bmatrix} e^{x} & xe^{x} & x^2e^{x} \\ e^{x} & e^{x}+xe^{x} & 2xe^{x} +x^2e^{x} \\ e^{x} & 2e^{x}+xe^{x} & 2e^{x}+4xe^{x}+x^2e^{x} \\ \end{bmatrix}$$
Below subtract row 1 from row 2 and subtract original row 2 from row 3:
$$W =\begin{bmatrix} e^{x} & xe^{x} & x^2e^{x} \\ 0 & e^{x} & 2xe^{x} \\ 0 & e^{x} & 2e^{x}+2xe^{x} \\ \end{bmatrix}$$
Below subtract new row 2 from row 3:
$$W =\begin{bmatrix} e^{x} & xe^{x} & x^2e^{x} \\ 0 & e^{x} & 2xe^{x} \\ 0 & 0 & 2e^{x} \\ \end{bmatrix} = e^{x}\begin{bmatrix} e^{x} & 2xe^{x} \\ 0 & 2e^{x} \\ \end{bmatrix} = 2e^{3x}$$
Indeed it appears that linear independence has been verified for all $x$ in $I$. I think?
Now let us superposition all functions and take their derivatives:
$$y_1=e^{x}; y_2=xe^{x}; y_3=x^2e^{x}$$
$$y= e^{x}+xe^{x}+x^2e^{x}$$
$$y'= 2e^{x}+3xe^{x}+x^2e^{x}$$
$$y''= 5e^{x}+5xe^{x}+x^2e^{x}$$
$$y'''= 10e^{x}+7xe^{x}+x^2e^{x}$$
We are now ready to substitute functions to establish basis in our given differential equation:
$$y'''-3y''+3y'-y=0$$
$$(10e^{x}+7xe^{x}+x^2e^{x})-3(5e^{x}+5xe^{x}+x^2e^{x})+3(2e^{x}+3xe^{x}+x^2e^{x}) -(e^{x}+xe^{x}+x^2e^{x})=0$$
$$0*e^{x} + 0*xe^{x} + 0*x^2e^{x}=0$$
We are now at the home stretch. Create an open interval $I=[-1,1]$ and select a few points $x_o$ and insert into $k_1*e^x+k_2xe^x+k_3*x^2e^x=0$:
When $x_o =-1$:
$$k_1*\frac{1}{e}-k_2\frac{1}{e}+k_3*\frac{1}{e}=k_1-k_2+k_3=0$$
Since this is true if we set all $k_i$ to zero we have established linear independence at this point within I.
When $x_o =0$:
$$k_1*e^0+k_2*0*e^0+k_3*0^2*e^0=k_1+0+0=0$$
Although this can be true when we only set $k_1$ to zero we still have the option of setting $k_2$ and $k_3$ to zero hence we have established linear independence at this point within I.
When $x_o =+1$:
$$k_1*e^1+k_2*1*e^1+k_3*1^2*e^1=k_1+k_2+k_3=0$$
We have now established linear independence for all three points within I.
This appears to be the solution. I think?