Given two vectors $x = \begin{pmatrix}t\\1\end{pmatrix}$ and $y = \begin{pmatrix}t^2\\2t\end{pmatrix}$ for $t > 0$. Find a system of homogeneous equations satisfied by these two vector functions.
What does this mean?
The first step of the problem was to check for linear dependence. I took the determinant so that $2t^2 - t^2 = t^2$. Since $t^2$ cannot equal zero given that $t > 0$, then the set of vectors are linearly independent.
What does it mean after to find a system of homogenous equations that satisfies these two vector functions?
Should I be finding eigenvalues and eigenvectors in terms of $t$ and $r$?
Thank you for the guidance.
You aren't asked to find a system of equations satisfying these two vector functions--what would it mean to satisfy a function? Rather, you're asked to find a system of equations satisfied by these two functions, meaning that these functions are solutions to said system.
For the sake of simplicity, let's start by looking for a system of first-order linear homogeneous equations to do the job. That is, we're looking for a system of the form: $$\begin{cases}v_1'+a_{1,1}(t)v_1+a_{1,2}(t)v_2=0\\v_2'+a_{2,1}(t)v_1+a_{2,2}(t)v_2=0\end{cases}\tag{$\star$}$$
What we need, then, is for $(\star)$ to hold when $(v_1,v_2)=(t,1)$ and when $(v_1,v_2)=(t^2,2t).$
Subbing $(v_1,v_2)=(t,1)$ into $(\star)$ gives $$\begin{cases}1+a_{1,1}(t)t+a_{1,2}(t)=0\\a_{2,1}(t)t+a_{2,2}(t)=0,\end{cases}$$ and subbing in $(t^2,2t)$ gives $$\begin{cases}2t+a_{1,1}(t)t^2+2a_{1,2}(t)t=0\\2+a_{2,1}(t)t^2+2a_{2,2}(t)t=0.\end{cases}$$ This gives us a system of $4$ equations to solve, which is good, because we have $4$ unknowns $a_{i,j}:=a_{i,j}(t)$ to figure out. Our system is: $$\begin{cases}1+a_{1,1}t+a_{1,2}=0\\a_{2,1}t+a_{2,2}=0\\2t+a_{1,1}t^2+2a_{1,2}t=0\\2+a_{2,1}t^2+2a_{2,2}t=0\end{cases}$$
By the second of these equations, we can substitute $$a_{2,2}=-a_{2,1}t\tag{1}$$ into the rest of our system, which then yields: $$\begin{cases}1+a_{1,1}t+a_{1,2}=0\\2t+a_{1,1}t^2+2a_{1,2}t=0\\2-a_{2,1}t^2=0\end{cases}$$
By the last of these equations together with $(1),$ we require $$a_{2,1}(t):=\frac2{t^2},a_{2,2}(t):=\frac{-2}t\tag{2}$$
On the other hand, we clearly require $$a_{1,2}=-1-a_{1,1}t\tag{3}$$ from the first equation of our starting system. Substituting into the third equation then yields $$-a_{1,1}t^2=0,$$ so by this equation and $(3)$ we must have $$a_{1,1}(t):=0,a_{1,2}(t)=-1.\tag{4}$$
At this point, we have that $(\star)$ holds $(v_1,v_2)=(t,1)$ and when $(v_1,v_2)=(t^2,2t),$ as desired.
There's another way we could go about this, though. Observe that your two vectors are both of the form $\bigl(f(t),f'(t)\bigr).$ So, one obvious homogeneous equation that they both must satisfy is $v_1'-v_2=0,$ which is exactly $v_1'+a_{1,1}v_1+a_{1,2}v_2=0,$ though with less work necessary.
At that point, we just need an equation of the form $$v_2'+c_1(t)v_1+c_2(t)v_2=0.$$ Subbing our desired solution vectors in gives us the system $$\begin{cases}0+c_1t+2c_2t=0\\2+c_1t^2+2c_2t=0,\end{cases}$$ which is readily solved to get $c_1=a_{2,1}$ and $c_2=a_{2,2}$ as above.