I started working on differential equations last week, and i had an exercise where i had to find two solutions for a differential equation(with different initial conditions). This is the equation: $$y''-3y' + 2y=0$$
So i have found the general solution for this equation, which is : $$ y= \alpha e^{2x} + \beta e^x$$
But the exercise asks me to find $y_1, y_2$ for these different initial conditions:
$\begin{cases} y_1(0)=1 \\ y_1'(0)=0 \end{cases}$ $\begin{cases} y_2(0)=0 & \\ y_2'(0)=1 & \end{cases}$
I found both solutions after solving two different systems of linear equations and this was the result:
$y_1 =e^{2x} - 2e^x$, $y_2=-e^{2x} + e^x$.
The problem comes when the exercise tells me i have to prove these equations $y_1$, $y_2$ are linearly independent in the vectorial space of all real functions $f:\mathbb{R}\longrightarrow\mathbb{R}$
I tried using knowledge i learnt from Linear Algebra:
$y_1,y_2$ are linearly independant if there exist scalars $a_1,a_2$ not all zero that: $a_1y_1 + a_2y_2=0$.
Thus, $$a_1(e^{2x} - 2e^x) + a_2(e^{2x} + e^x)=0$$
We can re-write this as: $$e^{2x}(a_1 + a_2) + e^x(-2a_1 + a_2)=0$$
if we solve this system : $\begin{cases} a_1 + a_2=0 \\ -2a_1 + a_2=0 \end{cases}$
We can deduce: $a_1 =0=a_2$. Then our functions are linearly independant, and we finished. Was my reasoning right? Is there any mistake?