General and particular solution for a matrix

Question

I'm given the matrix A and a vector b:

\begin{equation} A= \begin{pmatrix} 9 &-12 \\ 6 & -9 \end{pmatrix},\; b= \binom{7}{5} \end{equation} I have calculated the two eigenvalues and eigenvectors as:

\begin{equation} \lambda_{1}=3,\; \lambda_{2}=-3 \; \; and \; \; v_{1}= \binom{2}{1},\; v_{2}= \binom{1}{1} \end{equation}

The I found the general solution to A by: \begin{equation} x'(t)=Ax(t) \end{equation}

This gives me: \begin{equation} x=c_{1}\binom{2}{1}e^{3t}+c_{2}\binom{1}{1}e^{-3t} \end{equation}

Is that correct so far?

Then I have to give the particular solution that satisfies:

\begin{equation} x(0)=b \end{equation}

And now I'm stuck as I don't know how to proceed (b is the vector given above).

Thanks

Answer 1

$$x(0)=c_1\binom{2}{1}+c_2\binom{1}{1}=\binom{2c_1+c_2}{c_1+c_2}=\binom{7}{5}$$ $$\therefore c_1=2, c_2=3$$