$$\DeclareMathOperator{\abs}{abs}$$

We're working to show that the general solution of a linear, constant-coefficient differential equation can be given as the linear combination of solutions if these solution functions are linearly independent in the sense of nonzero Wronskian determinant.
For the 2nd order equation ay''+by'+cy=0, if r and s are distinct numbers and e^rx and e^sx are solutions, we claim that y=Ce^rx + De^rx is the general solution. Here's the Wronskian determinant for the distinct root case:

 (%i8) W(x):=matrix([%e^(r*x), %e^(s*x)],[diff(%e^(r*x),x), diff(%e^(s*x),x)]);
$\mathrm{\tt (\%o8) }\quad \mathrm{W}\left( x\right) :=\begin{pmatrix}{{\%e}^{r\cdot x}} & {{\%e}^{s\cdot x}}\cr \mathrm{diff}\left( {{\%e}^{r\cdot x}},x\right) & \mathrm{diff}\left( {{\%e}^{s\cdot x}},x\right) \end{pmatrix}$
 (%i9) determinant(W(x));
$\mathrm{\tt (\%o9) }\quad s\cdot {{\%e}^{s\cdot x+r\cdot x}}-r\cdot {{\%e}^{s\cdot x+r\cdot x}}$
 (%i10) factor(%);
$\mathrm{\tt (\%o10) }\quad \left( s-r\right) \cdot {{\%e}^{s\cdot x+r\cdot x}}$

and if we assume s and r are distinct, then this can't be zero.

Here's the Wronskian determinant for the repeated root case, where we claim the general solution is y=Ce^rx + Dxe^rx:

 --> W(x):=matrix([%e^(r*x), x*%e^(r*x)],[diff(%e^(r*x),x), diff(x*%e^(r*x),x)]);
$\mathrm{\tt (\%o5) }\quad \mathrm{W}\left( x\right) :=\begin{pmatrix}{{\%e}^{r\cdot x}} & x\cdot {{\%e}^{r\cdot x}}\cr \mathrm{diff}\left( {{\%e}^{r\cdot x}},x\right) & \mathrm{diff}\left( x\cdot {{\%e}^{r\cdot x}},x\right) \end{pmatrix}$
 (%i6) determinant(W(x));
$\mathrm{\tt (\%o6) }\quad {{\%e}^{r\cdot x}}\cdot \left( r\cdot x\cdot {{\%e}^{r\cdot x}}+{{\%e}^{r\cdot x}}\right) -r\cdot x\cdot {{\%e}^{2\cdot r\cdot x}}$
 (%i7) expand(%);
$\mathrm{\tt (\%o7) }\quad {{\%e}^{2\cdot r\cdot x}}$

which is never zero.

Now let's look at solutions of the 3rd order linear differential equation with 3 repeated roots:  (D-r)(D-r)(D-r)y=0 We claim the solutions are e^rx, xe^rx, and x^2e^rx Let's verify that they are solutions, and then Let's show that these three are linearly independent with a 3x3 Wronskian determinant=0

 (%i14) expand( (D-r)*(D-r)*(D-r));
$\mathrm{\tt (\%o14) }\quad {{D}^{3}}-3\cdot r\cdot {{D}^{2}}+3\cdot {{r}^{2}}\cdot D-{{r}^{3}}$

That showed us that the differential equation is y'''-3ry''+3r^2y'-r^3y Let's plug the solutions in and see what happens

 (%i16) y(x):=%e^(r*x);
$\mathrm{\tt (\%o16) }\quad \mathrm{y}\left( x\right) :={{\%e}^{r\cdot x}}$
 (%i21) diff(y(x),x,3)-3*r*diff(y(x),x,2)+3*r^2*diff(y(x),x)-r^3*y(x);
$\mathrm{\tt (\%o21) }\quad 0$
 (%i22) y(x):=x*%e^(r*x);
$\mathrm{\tt (\%o22) }\quad \mathrm{y}\left( x\right) :=x\cdot {{\%e}^{r\cdot x}}$
 (%i23) diff(y(x),x,3)-3*r*diff(y(x),x,2)+3*r^2*diff(y(x),x)-r^3*y(x);
$\mathrm{\tt (\%o23) }\quad -3\cdot r\cdot \left( {{r}^{2}}\cdot x\cdot {{\%e}^{r\cdot x}}+2\cdot r\cdot {{\%e}^{r\cdot x}}\right) +3\cdot {{r}^{2}}\cdot \left( r\cdot x\cdot {{\%e}^{r\cdot x}}+{{\%e}^{r\cdot x}}\right) +3\cdot {{r}^{2}}\cdot {{\%e}^{r\cdot x}}$
 (%i24) expand(%);
$\mathrm{\tt (\%o24) }\quad 0$
 (%i25) y(x):=x^2*%e^(r*x);
$\mathrm{\tt (\%o25) }\quad \mathrm{y}\left( x\right) :={{x}^{2}}\cdot {{\%e}^{r\cdot x}}$
 (%i26) diff(y(x),x,3)-3*r*diff(y(x),x,2)+3*r^2*diff(y(x),x)-r^3*y(x);
$\mathrm{\tt (\%o26) }\quad -3\cdot r\cdot \left( {{r}^{2}}\cdot {{x}^{2}}\cdot {{\%e}^{r\cdot x}}+4\cdot r\cdot x\cdot {{\%e}^{r\cdot x}}+2\cdot {{\%e}^{r\cdot x}}\right) +3\cdot {{r}^{2}}\cdot \left( r\cdot {{x}^{2}}\cdot {{\%e}^{r\cdot x}}+2\cdot x\cdot {{\%e}^{r\cdot x}}\right) +6\cdot {{r}^{2}}\cdot x\cdot {{\%e}^{r\cdot x}}+6\cdot r\cdot {{\%e}^{r\cdot x}}$
 (%i27) expand(%);
$\mathrm{\tt (\%o27) }\quad 0$

Boom!  and there they are, three solutions.  Now lets check linear independence in the sense of Wronskian Determinants:

 (%i11) W(x):=matrix([%e^(r*x), x*%e^(r*x),x^2*%e^(r*x)],    [diff(%e^(r*x),x), diff(x*%e^(r*x),x),diff(x^2*%e^(r*x),x)],    [diff(%e^(r*x),x,2), diff(x*%e^(r*x),x,2),diff(x^2*%e^(r*x),x,2)]);
$\mathrm{\tt (\%o11) }\quad \mathrm{W}\left( x\right) :=\begin{pmatrix}{{\%e}^{r\cdot x}} & x\cdot {{\%e}^{r\cdot x}} & {{x}^{2}}\cdot {{\%e}^{r\cdot x}}\cr \mathrm{diff}\left( {{\%e}^{r\cdot x}},x\right) & \mathrm{diff}\left( x\cdot {{\%e}^{r\cdot x}},x\right) & \mathrm{diff}\left( {{x}^{2}}\cdot {{\%e}^{r\cdot x}},x\right) \cr \mathrm{diff}\left( {{\%e}^{r\cdot x}},x,2\right) & \mathrm{diff}\left( x\cdot {{\%e}^{r\cdot x}},x,2\right) & \mathrm{diff}\left( {{x}^{2}}\cdot {{\%e}^{r\cdot x}},x,2\right) \end{pmatrix}$
 (%i12) determinant(W(x));
$\mathrm{\tt (\%o12) }\quad {{\%e}^{r\cdot x}}\cdot \left( \left( r\cdot x\cdot {{\%e}^{r\cdot x}}+{{\%e}^{r\cdot x}}\right) \cdot \left( {{r}^{2}}\cdot {{x}^{2}}\cdot {{\%e}^{r\cdot x}}+4\cdot r\cdot x\cdot {{\%e}^{r\cdot x}}+2\cdot {{\%e}^{r\cdot x}}\right) -\left( {{r}^{2}}\cdot x\cdot {{\%e}^{r\cdot x}}+2\cdot r\cdot {{\%e}^{r\cdot x}}\right) \cdot \left( r\cdot {{x}^{2}}\cdot {{\%e}^{r\cdot x}}+2\cdot x\cdot {{\%e}^{r\cdot x}}\right) \right) -x\cdot {{\%e}^{r\cdot x}}\cdot \left( r\cdot {{\%e}^{r\cdot x}}\cdot \left( {{r}^{2}}\cdot {{x}^{2}}\cdot {{\%e}^{r\cdot x}}+4\cdot r\cdot x\cdot {{\%e}^{r\cdot x}}+2\cdot {{\%e}^{r\cdot x}}\right) -{{r}^{2}}\cdot {{\%e}^{r\cdot x}}\cdot \left( r\cdot {{x}^{2}}\cdot {{\%e}^{r\cdot x}}+2\cdot x\cdot {{\%e}^{r\cdot x}}\right) \right) +{{x}^{2}}\cdot {{\%e}^{r\cdot x}}\cdot \left( r\cdot {{\%e}^{r\cdot x}}\cdot \left( {{r}^{2}}\cdot x\cdot {{\%e}^{r\cdot x}}+2\cdot r\cdot {{\%e}^{r\cdot x}}\right) -{{r}^{2}}\cdot {{\%e}^{r\cdot x}}\cdot \left( r\cdot x\cdot {{\%e}^{r\cdot x}}+{{\%e}^{r\cdot x}}\right) \right)$
 (%i13) expand(%);
$\mathrm{\tt (\%o13) }\quad 2\cdot {{\%e}^{3\cdot r\cdot x}}$

...and because this can't be zeros, the three functions are linearly independent

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