Consider a system of two second-order linear ODEs for which I have found a solution:
$$ \pmatrix{y_1 \\ y_2} = (A_1\cos\omega t + B_1\sin\omega t )\pmatrix{a \\ b} + \pmatrix{C_1 \\ C_2} $$
the constant term being due to an eigenvalue of zero for the matrix $A$ in the equation:
$$ \ddot{\boldsymbol{y}} = -A\cdot \boldsymbol{y} $$
Say I am now given an initial condition of $\dot{\boldsymbol{y}} = (v, 0)^T$ at $t = 0$. Differentiating my solution, I find:
$$ \dot{\boldsymbol{y}} = \pmatrix{v \\ 0} = B_1\omega\pmatrix{a \\ b} $$
Clearly there is no constant $B_1$ which satisfies this condition. How is it physically possible that my system cannot have an initial velocity? I'm really struggling to see what I might be doing wrong here, and can provide more context for how I arrived at my original equation if necessary. Sorry if this is confusing.
Original ODE:
$$ M\cdot\ddot{\boldsymbol{y}} = -\begin{bmatrix} k & -k \\ -k & k \end{bmatrix}\boldsymbol{y} $$
where $M = \mathrm{diag}(m_1, m_2)$, two known constants. $k$ is also known.