Unique solution to homogenous second order differential equation

I have a problem which gives me the homogenous second order differential equation $$y''-6y'+9y=0 \tag{1}\label{D.E.}$$ and had me find the expression which describes the solution to the differential equation (\ref{D.E.}). I found the solution to be: $$y(t)=c_1 e^{3t}+c_2 te^{3t} \tag{2}\label{solution}$$

The problem continues by stating:

The differential equation has a unique solution $y(t)$ with initial conditions $y(1)=3e^3, y'(1)=10e^3$. Mark the value $y(0)$ of that solution at $t=0$.

With the possible multiple choice answers: a) 6, b) 2, c) $0$ , and d) None of these.

I know how to solve for the IVP, but it throws me off that I have to solve for a unique solution (especially when the class and I haven't recieved any litterature on how to solve for a unique solution).

CALCULATIONS BELOW HAVE BEEN EDITED: (to correct $y'(t)$ this time)

I have tried solving for the IVP for $y(1)$ and $y'(1)$ where: $$y'(t)=3c_1e^{3t}+c_2\left(e^{3t}+3e^{3t}t\right) \tag{3}\label{y'solution}$$

And if I plug in $y(1)=3e^3$ in equation (\ref{solution}), and $y'(1)=10e^3$ in equation (\ref{y'solution}) then:

$$y(1)=c_1 e^{3\cdot1}+c_2\cdot1\cdot e^{3\cdot1}= 3e^3 \Rightarrow c_1 e^{3}+c_2e^{3}= 3e^3 \tag{4}\label{y(1)solution}$$

$$y'(1)=c_1e^{3\cdot1}\cdot 3+c_2\left(e^{3\cdot1}+3e^{3\cdot1}\cdot1\right) = 10e^3 \Rightarrow 3c_1e^{3}+c_2\left(e^{3}+3e^{3}\right) = 10e^3 \tag{5}\label{y'(1)solution}$$

If we continue, and divide both sides of $y(1)$ with $e^3$ we get: $$y(1)=\frac{(c_1 e^3+c_2e^3)}{e^3}=\frac{3e^3}{e^3} \Rightarrow c_1+c_2=3$$

We then multiply the $c_2(e^{3}+3e^{3})$ part out and do the same:

$$y'(1)=\frac{(3c_1e^{3}+c_2e^{3}+3c_2e^{3})}{e^3} = \frac{10e^3}{e^3} \Rightarrow 3c_1+4c_2=10$$

and get:

$$\begin{cases} c_1 + c_2 = 3 \\ 3c_1 + 4c_2 = 10 \end{cases}$$

which yields:

$y(t) = 2e^{3t} + te^{3t}\tag{6}\label{y(t)}$

Finishing it of with $y(0)$ at $t=0$ we get: $$y(0) = 2e^{3\cdot0} + 0\cdot e^{3\cdot0} = 2$$
You differentiate incorrectly. The IVP system should look like this: $$\begin{cases} c_1 + c_2 = 3 \\ 3c_1 + 4c_2 = 10 \end{cases}$$ yielding the following answer $y(t) = 2e^{3t} + te^{3t}$ and the correct answer is b).
Your derivative in $(3)$ is wrong ! You have to use the product rule for the derivative of $te^t$ !