Euler-Lagrange equations solving for Extreme points of functional

Question

I'm trying to solve the Euler-Langrange equation for the functional $$F(t, (x,u,p), (\mathring{x}, \mathring{u}, \mathring{p}))= \dfrac{1}{2}[x(t)^{T}Qx(t)+u(t)^{T}Ru(t)]+p(t)^{T}[\mathring{x}(t)-Ax(t)-Bu(t)]$$ where Q,R are symmetric matrices, R is possitive definite, Q is semi-possitive definite. My work so far is: $$ \dfrac{\partial F}{\partial x}-\dfrac{d}{dt} \dfrac{\partial F}{\partial \mathring{x}}=0 \leftrightarrow \begin{bmatrix} Qx(t)-A^{T}p(t)-\mathring{p}(t)=0 (1) \\ Ru(t)-B^{T}p(t)=0 (2)\\ \mathring{x}(t)-Ax(t)-Bu(t)=0(3) \end{bmatrix} $$ (2) gives: $$u(t)=R^{-1}B^{T}p(t)$$ Replacing that in (3) we get: $$p(t)=(BR^{-1}B^{T})^{-1}\mathring{x}(t)-(BR^{-1}B^{T})^{-1}Ax(t)$$ and due to that $$\mathring{p}(t)=(BR^{-1}B^{T})^{-1}\ddot{x}(t)-(BR^{-1}B^{T})^{-1}A\mathring{x}(t)$$ so by replacing $p(t)$ and $\mathring{p}(t)$ in (1): $$(BR^{-1}B^{T})^{-1}\ddot{x}(t)+[A^{T}(BR^{-1}B^{T})^{-1}-(BR^{-1}B^{T})^{-1}A]\mathring{x}(t)-[A^{T}(BR^{-1}B^{T})^{-1}A+Q]x(t)=0$$ So how can I solve this differential equation?