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EDIT 1: I think one needs to assume that $\|B\|<2$, since otherwise as @Iosif Pinelis example in the comments shows, the eigenvalues may be purely imaginary.

EDIT 2: I changed the notation so my answer become clearer. In the original answer, I had the following fact in mind: if $P(\xi)$ is the principle symbol of a linear $2k$-differential operator satisfying $P(\xi)\geq c|\xi|^{2}$, then the eigenvalues of the $2m$-order linear ODE $P(\xi+i\frac{d}{dt}\nu)y(t)=0$ must have the desired properties in the question. I basically reconstructed such a symbol from the given ODE, but then I reckoned this will not be needed in the actual answer.

Under the assumption that $\|B\|<2$, let us see that all eigenvalues have either positive or negative real part. Consider the matrix-valued polynomial of degree 2 in $\lambda$, $P:\mathbb{R}\rightarrow\mathrm{End}{\mathbb(\mathbb{R}^{d})}$, given by $$P(\lambda)=\lambda^{2}A-\lambda B+C$$ Note how for every $\lambda\in\mathbb{R}$, by the assumptions in the question we find $$c(1+\lambda^{2})I\leq P(\lambda)$$ For some constant $c>0$. In particular, $P(\lambda)$ is nonsingular for all $\lambda\in\mathbb{R}$.

By formally allowing $\lambda$ to be complex, we note how the eigenvalue problem for the ODE in the question becomes for $y(t)=y_{0}e^{\lambda t}$, $$P(i\lambda)y_{0}=0$$ For $\lambda=i\eta$ with $\eta\in\mathbb{R}$, this reads, $$P(-\eta)y_{0}=0$$ which implies $y_{0}=0$. Therefore, the eigenvalues of the ODE can't be purely imaginary, and must contain a real part.

This implies that the fundamental solutions exponentially decreases when either $t\rightarrow\infty$ or $t\rightarrow-\infty$. Let $M^{+}$ denote the space of $t\rightarrow\infty$ exponentially decreasing solutions and by $M^{-}$ the space of $t\rightarrow-\infty$ exponentially decreasing solutions. Let $I:M^{\pm}\rightarrow \mathbb{R}^{d}$ be the map $Iy=y(0)$. This map is clearly linear. We prove that this map is injective, hence $\operatorname{dim}{M^{\pm}}\leq d$. Since the space of solutions is $2d$ dimensional, this implies that $\operatorname{dim}{M^{\pm}}=d$, which is the required result. The proof also shows that all fundamental solutions must be of the form $y(t)=y_{0}e^{\lambda t}$, because if $y(t)=y_{0}t^{k}e^{\lambda t}$ for some $k\ne 0$, then $y(0)=0$, hence $y=0$ identically.

Without loss of generality, let $y\in M^{+}$ with $y(0)=0$. Denote by $\hat{y}(\eta)$ the fourier transform of $y$, albeit extneded to satisfy $y(t)=0$ for $t\leq 0$. Since $P(\lambda)$ is positive definite for all $\lambda\in\mathbb{R}$, we may write: $P(\eta)=M(\eta)M(\eta)$, where $M(\eta)$ is a matrix-valued polynomial of degree at most 1 in $\eta$.

By formally replacing $\eta$ with an ordinary derivative $i\frac{d}{dt}$, the ODE can formally be written as, $$P(i\frac{d}{dt})y(t)=0$$ Since $y(0)=0$ and the solution is rapidly decreasing, we can integrate by parts the following equation to obtain, $$0=(P(i\frac{d}{dt})y,y)_{L^{2}(\mathbb{R^+})}=(M(i\frac{d}{dt})y,M(i\frac{d}{dt})y)_{L^{2}(\mathbb{R^+})}$$ By the Plancheral theorem, this quantity becomes by applying the Fourier transform on each flank, $$\int_{-\infty}^{\infty}(P(\eta)\hat{y}(\eta),\hat{y}(\eta))_{\mathbb{R}^{d}}d\eta=0$$.
And since $P(\eta)\geq c(1+\eta^{2})I$ for all $\eta\in\mathbb{R}$, this yields $\hat{y}(\eta)=0$, which is to say $y=0$. This proves that $I$ was injective, and the solution space decomposes as described.

EDIT 1: I think one needs to assume that $\|B\|<2$, since otherwise as @Iosif Pinelis example in the comments shows, the eigenvalues may be purely imaginary.

EDIT 2: I changed the notation so my answer become clearer. In the original answer, I had the following fact in mind: if $P(\xi)$ is the principle symbol of a linear $2$-differential operator satisfying $P(\xi)\geq c|\xi|^{2}$, then the eigenvalues of the $2$-order linear ODE $P(\xi+i\frac{d}{dt}\nu)y(t)=0$ must have the desired properties in the question. I basically reconstructed such a symbol from the given ODE, but then I reckoned this will not be needed in the actual answer.

Under the assumption that $\|B\|<2$, let us see that all eigenvalues have either positive or negative real part. Consider the matrix-valued polynomial of degree 2 in $\lambda$, $P:\mathbb{R}\rightarrow\mathrm{End}{\mathbb(\mathbb{R}^{d})}$, given by $$P(\lambda)=\lambda^{2}A-\lambda B+C$$ Note how for every $\lambda\in\mathbb{R}$, by the assumptions in the question we find $$c(1+\lambda^{2})I\leq P(\lambda)$$ For some constant $c>0$. In particular, $P(\lambda)$ is nonsingular for all $\lambda\in\mathbb{R}$.

By formally allowing $\lambda$ to be complex, we note how the eigenvalue problem for the ODE in the question becomes for $y(t)=y_{0}e^{\lambda t}$, $$P(i\lambda)y_{0}=0$$ For $\lambda=i\eta$ with $\eta\in\mathbb{R}$, this reads, $$P(-\eta)y_{0}=0$$ which implies $y_{0}=0$. Therefore, the eigenvalues of the ODE can't be purely imaginary, and must contain a real part.

This implies that the fundamental solutions exponentially decreases when either $t\rightarrow\infty$ or $t\rightarrow-\infty$. Let $M^{+}$ denote the space of $t\rightarrow\infty$ exponentially decreasing solutions and by $M^{-}$ the space of $t\rightarrow-\infty$ exponentially decreasing solutions. Let $I:M^{\pm}\rightarrow \mathbb{R}^{d}$ be the map $Iy=y(0)$. This map is clearly linear. We prove that this map is injective, hence $\operatorname{dim}{M^{\pm}}\leq d$. Since the space of solutions is $2d$ dimensional, this implies that $\operatorname{dim}{M^{\pm}}=d$, which is the required result. The proof also shows that all fundamental solutions must be of the form $y(t)=y_{0}e^{\lambda t}$, because if $y(t)=y_{0}t^{k}e^{\lambda t}$ for some $k\ne 0$, then $y(0)=0$, hence $y=0$ identically.

Without loss of generality, let $y\in M^{+}$ with $y(0)=0$. Denote by $\hat{y}(\eta)$ the fourier transform of $y$, albeit extneded to satisfy $y(t)=0$ for $t\leq 0$. Since $P(\lambda)$ is positive definite for all $\lambda\in\mathbb{R}$, we may write: $P(\eta)=M(\eta)M(\eta)$, where $M(\eta)$ is a matrix-valued polynomial of degree at most 1 in $\eta$.

By formally replacing $\eta$ with an ordinary derivative $i\frac{d}{dt}$, the ODE can formally be written as, $$P(i\frac{d}{dt})y(t)=0$$ Since $y(0)=0$ and the solution is rapidly decreasing, we can integrate by parts the following equation to obtain, $$0=(P(i\frac{d}{dt})y,y)_{L^{2}(\mathbb{R^+})}=(M(i\frac{d}{dt})y,M(i\frac{d}{dt})y)_{L^{2}(\mathbb{R^+})}$$ By the Plancheral theorem, this quantity becomes by applying the Fourier transform on each flank, $$\int_{-\infty}^{\infty}(P(\eta)\hat{y}(\eta),\hat{y}(\eta))_{\mathbb{R}^{d}}d\eta=0$$.
And since $P(\eta)\geq c(1+\eta^{2})I$ for all $\eta\in\mathbb{R}$, this yields $\hat{y}(\eta)=0$, which is to say $y=0$. This proves that $I$ was injective, and the solution space decomposes as described.

EDIT: I think one needs to assume that $\|B\|<2$, since otherwise as @Iosif Pinelis example in the comments shows, the eigenvalues may be purely imaginary.

Under the assumption that $\|B\|<2$, let us see that all eigenvalues have either positive or negative real part. Let $(\mu,\nu)$ be the standard basis for $\mathbb{R}^{2}$. Decompose every vector $\xi\in\mathbb{R}^{2}$ into $\xi=\xi_{\mu}\mu+\lambda\nu$. We consider the matrix-valued quadratic form $P:\mathbb{R}^{2}\rightarrow\mathrm{End}{\mathbb(\mathbb{R}^{d})}$ given by $$P(\xi_{\mu}\mu+\lambda\nu)=\lambda^{2}A-\lambda\xi_{\mu} B+\xi_{\mu}^{2}C$$ Note how for every $\xi\in\mathbb{R}^{2}$, by the assumptions in the question we find $$c|\xi|^{2}I\leq P(\xi)$$ For some constant $c>0$. In particular, $P(\xi)$ is nonsingular for all $\xi\ne 0$.

By formally allowing $\lambda,\xi_{\mu}$ to be complex, we note how the eigenvalue problem for the ODE in the question becomes for $y(t)=y_{0}e^{\lambda t}$, $$P(\mu+i\lambda\nu)y_{0}=0$$ For $\lambda=i\eta$ with $\eta\in\mathbb{R}$, this reads, $$P(\mu-\eta\nu)y_{0}=0$$ which implies $y_{0}=0$. Therefore, the eigenvalues of the ODE can't be purely imaginary, and must contain a real part.

This implies that the fundamental solutions exponentially decreases when either $t\rightarrow\infty$ or $t\rightarrow-\infty$. Let $M^{+}$ denote the space of $t\rightarrow\infty$ exponentially decreasing solutions and by $M^{-}$ the space of $t\rightarrow-\infty$ exponentially decreasing solutions. Let $I:M^{\pm}\rightarrow \mathbb{R}^{d}$ be the map $Iy=y(0)$. This map is clearly linear. We prove that this map is injective, hence $\operatorname{dim}{M^{\pm}}\leq d$. Since the space of solutions is $2d$ dimensional, this implies that $\operatorname{dim}{M^{\pm}}=d$, which is the required result.

Without loss of generality, let $y\in M^{+}$ with $y(0)=0$. Denote by $\hat{y}(\eta)$ the fourier transform of $y$, albeit extneded to satisfy $y(t)=0$ for $t\leq 0$. For all $\eta\in\mathbb{R}$, define $Q(\eta)$ to be the operator $$Q(\eta)=P(\mu-\eta\nu).$$

Decompose $Q(\eta)=M(\eta)L(\eta)$, where $L$ and $M$ are both linear in $\eta$.

By formally replacing $\eta$ with an ordinary derivative $i\frac{d}{dt}$, the ODE can formally be written as, $$Q(i\frac{d}{dt})y(t)=P(\mu-i\frac{d}{dt})y(t)=0$$ Since $y(0)=0$ and the solution is rapidly decreasing, we can integrate by parts the following equation to obtain, $$0=(Q(i\frac{d}{dt})y,y)_{L^{2}(\mathbb{R^+})}=(L(i\frac{d}{dt})y,M^{*}(i\frac{d}{dt})y)_{L^{2}(\mathbb{R^+})}$$ By the Plancheral theorem, this quantity becomes, $$\int_{-\infty}^{\infty}(P(\mu-\eta\nu)\hat{y}(\eta),\hat{y}(\eta))_{\mathbb{R}^{d}}d\eta=0$$.
And since $P(\xi)\geq c|\xi|^{2}I$ for all $\xi\ne 0$, this yields $\hat{y}(\eta)=0$, which is to say $y=0$. This proves that $I$ was injective, and the solution space decomposes as described.

The proof also shows that all fundamental solutions must be of the form $y(t)=y_{0}e^{\lambda t}$, because if $y(t)=y_{0}t^{k}e^{\lambda t}$ for some $k\ne 0$, then $y(0)=0$, hence $y=0$ identically.

EDIT 1: I think one needs to assume that $\|B\|<2$, since otherwise as @Iosif Pinelis example in the comments shows, the eigenvalues may be purely imaginary.

EDIT 2: I changed the notation so my answer become clearer. In the original answer, I had the following fact in mind: if $P(\xi)$ is the principle symbol of a linear $2k$-differential operator satisfying $P(\xi)\geq c|\xi|^{2}$, then the eigenvalues of the $2m$-order linear ODE $P(\xi+i\frac{d}{dt}\nu)y(t)=0$ must have the desired properties in the question. I basically reconstructed such a symbol from the given ODE, but then I reckoned this will not be needed in the actual answer.

Under the assumption that $\|B\|<2$, let us see that all eigenvalues have either positive or negative real part. Consider the matrix-valued polynomial of degree 2 in $\lambda$, $P:\mathbb{R}\rightarrow\mathrm{End}{\mathbb(\mathbb{R}^{d})}$, given by $$P(\lambda)=\lambda^{2}A-\lambda B+C$$ Note how for every $\lambda\in\mathbb{R}$, by the assumptions in the question we find $$c(1+\lambda^{2})I\leq P(\lambda)$$ For some constant $c>0$. In particular, $P(\lambda)$ is nonsingular for all $\lambda\in\mathbb{R}$.

By formally allowing $\lambda$ to be complex, we note how the eigenvalue problem for the ODE in the question becomes for $y(t)=y_{0}e^{\lambda t}$, $$P(i\lambda)y_{0}=0$$ For $\lambda=i\eta$ with $\eta\in\mathbb{R}$, this reads, $$P(-\eta)y_{0}=0$$ which implies $y_{0}=0$. Therefore, the eigenvalues of the ODE can't be purely imaginary, and must contain a real part.

This implies that the fundamental solutions exponentially decreases when either $t\rightarrow\infty$ or $t\rightarrow-\infty$. Let $M^{+}$ denote the space of $t\rightarrow\infty$ exponentially decreasing solutions and by $M^{-}$ the space of $t\rightarrow-\infty$ exponentially decreasing solutions. Let $I:M^{\pm}\rightarrow \mathbb{R}^{d}$ be the map $Iy=y(0)$. This map is clearly linear. We prove that this map is injective, hence $\operatorname{dim}{M^{\pm}}\leq d$. Since the space of solutions is $2d$ dimensional, this implies that $\operatorname{dim}{M^{\pm}}=d$, which is the required result. The proof also shows that all fundamental solutions must be of the form $y(t)=y_{0}e^{\lambda t}$, because if $y(t)=y_{0}t^{k}e^{\lambda t}$ for some $k\ne 0$, then $y(0)=0$, hence $y=0$ identically.

Without loss of generality, let $y\in M^{+}$ with $y(0)=0$. Denote by $\hat{y}(\eta)$ the fourier transform of $y$, albeit extneded to satisfy $y(t)=0$ for $t\leq 0$. Since $P(\lambda)$ is positive definite for all $\lambda\in\mathbb{R}$, we may write: $P(\eta)=M(\eta)M(\eta)$, where $M(\eta)$ is a matrix-valued polynomial of degree at most 1 in $\eta$.

By formally replacing $\eta$ with an ordinary derivative $i\frac{d}{dt}$, the ODE can formally be written as, $$P(i\frac{d}{dt})y(t)=0$$ Since $y(0)=0$ and the solution is rapidly decreasing, we can integrate by parts the following equation to obtain, $$0=(P(i\frac{d}{dt})y,y)_{L^{2}(\mathbb{R^+})}=(M(i\frac{d}{dt})y,M(i\frac{d}{dt})y)_{L^{2}(\mathbb{R^+})}$$ By the Plancheral theorem, this quantity becomes by applying the Fourier transform on each flank, $$\int_{-\infty}^{\infty}(P(\eta)\hat{y}(\eta),\hat{y}(\eta))_{\mathbb{R}^{d}}d\eta=0$$.
And since $P(\eta)\geq c(1+\eta^{2})I$ for all $\eta\in\mathbb{R}$, this yields $\hat{y}(\eta)=0$, which is to say $y=0$. This proves that $I$ was injective, and the solution space decomposes as described.

First, let us see that all eigenvalues have either positive or negative real part. Let $(\mu,\nu)$ be the standard basis for $\mathbb{R}^{2}$. Decompose every vector $\xi\in\mathbb{R}^{2}$ into $\xi=\xi_{\mu}\mu+\lambda\nu$. We consider the matrix-valued quadratic form $P:\mathbb{R}^{2}\rightarrow\mathrm{End}{\mathbb(\mathbb{R}^{d})}$ given by $$P(\xi_{\mu}\mu+\lambda\nu)=\lambda^{2}A-\lambda\xi_{\mu} B+\xi_{\mu}^{2}C$$ Note how for every $\xi\in\mathbb{R}^{2}$, by the assumptions in the question we find $$|\xi|^{2}I\leq P(\xi)\leq 2|\xi|^{2}I.$$ In particular, $P(\xi)$ is nonsingular for all $\xi\ne 0$.

By formally allowing $\lambda,\xi_{\mu}$ to be complex, we note how the eigenvalue problem for the ODE in the question becomes for $y(t)=y_{0}e^{\lambda t}$, $$P(\mu+i\lambda\nu)y_{0}=0$$ For $\lambda=i\eta$ with $\eta\in\mathbb{R}$, this reads, $$P(\mu-\eta\nu)y_{0}=0$$ which implies $y_{0}=0$. Therefore, the eigenvalues of the ODE can't be purely imaginary, and must contain a real part.

This implies that the fundamental solutions exponentially decreases when either $t\rightarrow\infty$ or $t\rightarrow-\infty$. Let $M^{+}$ denote the space of $t\rightarrow\infty$ exponentially decreasing solutions and by $M^{-}$ the space of $t\rightarrow-\infty$ exponentially decreasing solutions. Let $I:M^{\pm}\rightarrow \mathbb{R}^{d}$ be the map $Iy=y(0)$. This map is clearly linear. We prove that this map is injective, hence $\operatorname{dim}{M^{\pm}}\leq d$. Since the space of solutions is $2d$ dimensional, this implies that $\operatorname{dim}{M^{\pm}}=d$, which is the required result.

Without loss of generality, let $y\in M^{+}$ with $y(0)=0$. Denote by $\hat{y}(\eta)$ the fourier transform of $y$, albeit extneded to satisfy $y(t)=0$ for $t\leq 0$. For all $\eta\in\mathbb{R}$, define $Q(\eta)$ to be the operator $$Q(\eta)=P(\mu-\eta\nu).$$

Decompose $Q(\eta)=M(\eta)L(\eta)$, where $L$ and $M$ are both linear in $\eta$.

By formally replacing $\eta$ with an ordinary derivative $i\frac{d}{dt}$, the ODE can formally be written as, $$Q(i\frac{d}{dt})y(t)=P(\mu-i\frac{d}{dt})y(t)=0$$ Since $y(0)=0$ and the solution is rapidly decreasing, we can integrate by parts the following equation to obtain, $$0=(Q(i\frac{d}{dt})y,y)_{L^{2}(\mathbb{R^+})}=(L(i\frac{d}{dt})y,M^{*}(i\frac{d}{dt})y)_{L^{2}(\mathbb{R^+})}$$ By the Plancheral theorem, this quantity becomes, $$\int_{-\infty}^{\infty}(P(\mu-\eta\nu)\hat{y}(\eta),\hat{y}(\eta))_{\mathbb{R}^{d}}d\eta=0$$.
And since $P(\xi)\geq |\xi|^{2}I$ for all $\xi\ne 0$, this yields $\hat{y}(\eta)=0$, which is to say $y=0$. This proves that $I$ was injective, and the solution space decomposes as described.

The proof also shows that all fundamental solutions must be of the form $y(t)=y_{0}e^{\lambda t}$, because if $y(t)=y_{0}t^{k}e^{\lambda t}$ for some $k\ne 0$, then $y(0)=0$, hence $y=0$ identically.

EDIT: I think one needs to assume that $\|B\|<2$, since otherwise as @Iosif Pinelis example in the comments shows, the eigenvalues may be purely imaginary.

Under the assumption that $\|B\|<2$, let us see that all eigenvalues have either positive or negative real part. Let $(\mu,\nu)$ be the standard basis for $\mathbb{R}^{2}$. Decompose every vector $\xi\in\mathbb{R}^{2}$ into $\xi=\xi_{\mu}\mu+\lambda\nu$. We consider the matrix-valued quadratic form $P:\mathbb{R}^{2}\rightarrow\mathrm{End}{\mathbb(\mathbb{R}^{d})}$ given by $$P(\xi_{\mu}\mu+\lambda\nu)=\lambda^{2}A-\lambda\xi_{\mu} B+\xi_{\mu}^{2}C$$ Note how for every $\xi\in\mathbb{R}^{2}$, by the assumptions in the question we find $$c|\xi|^{2}I\leq P(\xi)$$ For some constant $c>0$. In particular, $P(\xi)$ is nonsingular for all $\xi\ne 0$.

By formally allowing $\lambda,\xi_{\mu}$ to be complex, we note how the eigenvalue problem for the ODE in the question becomes for $y(t)=y_{0}e^{\lambda t}$, $$P(\mu+i\lambda\nu)y_{0}=0$$ For $\lambda=i\eta$ with $\eta\in\mathbb{R}$, this reads, $$P(\mu-\eta\nu)y_{0}=0$$ which implies $y_{0}=0$. Therefore, the eigenvalues of the ODE can't be purely imaginary, and must contain a real part.

This implies that the fundamental solutions exponentially decreases when either $t\rightarrow\infty$ or $t\rightarrow-\infty$. Let $M^{+}$ denote the space of $t\rightarrow\infty$ exponentially decreasing solutions and by $M^{-}$ the space of $t\rightarrow-\infty$ exponentially decreasing solutions. Let $I:M^{\pm}\rightarrow \mathbb{R}^{d}$ be the map $Iy=y(0)$. This map is clearly linear. We prove that this map is injective, hence $\operatorname{dim}{M^{\pm}}\leq d$. Since the space of solutions is $2d$ dimensional, this implies that $\operatorname{dim}{M^{\pm}}=d$, which is the required result.

Without loss of generality, let $y\in M^{+}$ with $y(0)=0$. Denote by $\hat{y}(\eta)$ the fourier transform of $y$, albeit extneded to satisfy $y(t)=0$ for $t\leq 0$. For all $\eta\in\mathbb{R}$, define $Q(\eta)$ to be the operator $$Q(\eta)=P(\mu-\eta\nu).$$

Decompose $Q(\eta)=M(\eta)L(\eta)$, where $L$ and $M$ are both linear in $\eta$.

By formally replacing $\eta$ with an ordinary derivative $i\frac{d}{dt}$, the ODE can formally be written as, $$Q(i\frac{d}{dt})y(t)=P(\mu-i\frac{d}{dt})y(t)=0$$ Since $y(0)=0$ and the solution is rapidly decreasing, we can integrate by parts the following equation to obtain, $$0=(Q(i\frac{d}{dt})y,y)_{L^{2}(\mathbb{R^+})}=(L(i\frac{d}{dt})y,M^{*}(i\frac{d}{dt})y)_{L^{2}(\mathbb{R^+})}$$ By the Plancheral theorem, this quantity becomes, $$\int_{-\infty}^{\infty}(P(\mu-\eta\nu)\hat{y}(\eta),\hat{y}(\eta))_{\mathbb{R}^{d}}d\eta=0$$.
And since $P(\xi)\geq c|\xi|^{2}I$ for all $\xi\ne 0$, this yields $\hat{y}(\eta)=0$, which is to say $y=0$. This proves that $I$ was injective, and the solution space decomposes as described.

The proof also shows that all fundamental solutions must be of the form $y(t)=y_{0}e^{\lambda t}$, because if $y(t)=y_{0}t^{k}e^{\lambda t}$ for some $k\ne 0$, then $y(0)=0$, hence $y=0$ identically.