Solving a system of ODEs with a zero eigenvalue and non-zero initial velocity

Question

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.

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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.

Answer 1

As $\left( \begin{array}{cc} k & -k \\ -k & k \\ \end{array} \right) = T^{-1}\Lambda T$ with $T = \left( \begin{array}{cc} -1 & 1 \\ 1 & 1 \\ \end{array} \right)$ and $\Lambda = \left( \begin{array}{cc} 2 k & 0 \\ 0 & 0 \\ \end{array} \right)$ we have

$$ m\ddot y + T^{-1}\Lambda Ty=0\ \ \Rightarrow mT\ddot y + \Lambda T y = 0 $$

so calling $z = T y$ we have the decoupled linear system

$$ m\ddot z + \Lambda z = 0 $$

with solution

$$ \cases{ z_1 = c_1\cos\left(\sqrt{\frac{2k}{m}}t\right)+ c_2\sin\left(\sqrt{\frac{2k}{m}}t\right)\\ z_2 = c_3+c_4 t } $$

so we have four independent constants to configure with the initial/boundary conditions. From $z$ to $y$ we proceed as $y = T^{-1}z$. Concluding, we are free to configure those initial/boundary conditions as needed.