When solving differential equations, does the solution order matters? Example with a repeated root solution: is $\mathbf{y}_1=\mathbf{C}_1.e^{x/2}.x ,$$ \mathbf{y}_2 = \mathbf{C}_2e^{x/2}$ the same thing as $\mathbf{y}_2 =\mathbf{C}_1.e^{x/2}.x,\mathbf{y}_1=\mathbf{C}_2.e^{x/2}$? Or more generally,when you're solving second order differential equations,when does the order of the solutions matter at all? Thanks!
 A: As Andrew mentions, it is completely arbitrary. 
Lets reverse engineer your problem and show this. We have:
$$\tag 1 y''(x)-y'(x)+ \dfrac{1}{4}y(x) = 0$$
The roots of CP are given by:
$$m^2 - m + \dfrac{1}{4} = 0 \rightarrow m_{1,2} = \dfrac{1}{2}$$
So, we have a double root, thus we can choose solutions as:
$$\tag 2 y_1 = c_1 e^{x/2}+c_2 e^{x/2} x$$
or
$$\tag 3 y_2 = +c_1 e^{x/2} x + c_2 e^{x/2}$$
or other variants with negative constants.
Lets verify these two variants of these solutions.
Solution 1:


*

*$y_1 = c_1 e^{x/2}+c_2 e^{x/2} x$

*$y'_1 = \dfrac{1}{2} c_1e^{x/2} + c_2e^{x/2}\left(\dfrac{x}{2} + 1\right)$

*$y''_1 = \dfrac{1}{4} e^{x/2} \left(c_2 (x+4)+c_1 \right)$

*Substituting back into $(1)$ yields:
$y_1''-y_1'+ \dfrac{1}{4}y_1 = \left(\dfrac{1}{4} e^{x/2} (c_2 (x+4)+c_1)\right) -\left(\dfrac{1}{2} c_1e^{x/2} + c_2e^{x/2}\left(\dfrac{x}{2} + 1\right)\right) + \dfrac{1}{4}\left(c_1 e^{x/2}+c_2 e^{x/2} x\right) = 0$
So, solution $1$ works. Repeat this for $(3)$, that is $y_2$, and verify that it also solves the DEQ to convince yourself.
