It is a particular case of the linear complex case and, more in general, of the system of first order ODEs.
If $A$ is a real matrix, given a solution $\varphi(t)$, the conjugate function is just the solution of the system whose initial value is the conjugate of that of $\varphi(t)$.
By uniqueness of solutions, if we take any real initial condition, $v\in \mathbb{R}^n$ then
$$ e^{At}\cdot v $$must be real, and so $e^{At}$ is a real matrix.
What form does the elements of this matrix have? In the note of the complex case we stated that
$$ e^{At}=M e^{Jt} M^{-1} $$Solutions are then made by mixing quasi-polynomials by mean of complex number, so the results are linear combinations of real and imaginary parts of these quasi-polinomials:
Let $\dot{z}=Az$ a system of first order linear equations, with $A:\mathbb{R}^n \mapsto \mathbb{R}^n$ a linear operator. Then, if $A$ has real eigenvalues $\lambda_i$ with algebraic multiplicities $n_i$ and complex eigenvalues $\alpha_j\pm i \omega_j$ with multiplicities $\mu_j$ then every component of a solution $\varphi(t)$ has the form:
$$ \varphi_{m}(t)= \sum_{i} e^{\lambda_{i} t} p_{m, i}+ \sum_{j} e^{\alpha_{j} t}\left[q_{m, j}(t) \cos \omega_{j} t+r_{m, j}(t) \sin \omega_{j} t\right] $$where $p_{m,i}(t)$, $q_{m,j}(t)$ and $r_{m,j}(t)$ are real polynomials with degree limited by the multiplicities.
$\blacksquare$
In particular, consider the case where \textbf{all eigenvalues are single}. We would have:
$$ \varphi_{m}(t)= \sum_{i} c_{m, i} e^{\lambda_{i} t} + \sum_{j} a_{m, j} e^{\alpha_{j} t} \cos \omega_{j} t+b_{m, j} e^{\alpha_{j} t}\sin \omega_{j} t $$(1)
with $a, b, c$ families of real constants. But there are too many.
How many of degrees of freedom do we have? Let's see.
The map $\varphi:I\subset \mathbb{R}\mapsto \mathbb{C}^n$ is a complex solution of the equation if and only if the real and the imaginary part of $z$ are both real solutions. This is because of linearity and because $\overline{z}(t)$ is also a solution of the associated complex equation (the only one with initial value the complex conjugate of $z(0)$).
Suppose $A$ has $n$ distinct eigenvalues, $r$ real eigenvalues and $c$ complex eigenvalues, such that $n=r+2c$. The general solution is
$$\varphi(t)=\sum_{k=1}^{n} c_{k} e^{\lambda_{k} t} \xi_{k}$$
where we have chosen real eigenvectors for the real eigenvalues, and complex conjugate eigenvectors for the complex eigenvalue pairs. We observe that for $\varphi(0)$ being real
$$\sum_{k=1}^{n} c_{k} \boldsymbol{\xi}_{k}=\sum_{k=1}^{n} \overline{c}_{k} \overline{\boldsymbol{\xi}}_{k}$$
and therefore coefficients of complex conjugates eigenvectors must be complex conjugates. So, for real eigenvalues (with real eigenvector) $c_k$ is real, and we denote it by $a_k$. And for the complex conjugates eigenvalues $\lambda_k$ and $\lambda_j$, with of course complex conjugates eigenvectors, coefficients $c_k$ and $c_j$ must be conjugates. So we can write the solution as
$$ \varphi(t)=\sum_{k=1}^{r} a_{k} e^{\lambda_{k} t} \xi_{k}+\sum_{k=r+1}^{r+c} c_{k} e^{\lambda_{k} t} \xi_{k}+\sum_{k=r+1}^{r+c} \overline{c_{k}} e^{\overline{\lambda_{k}} t} \overline{\xi_{k}} $$and so
$$ \varphi(t)=\sum_{k=1}^{r} a_{k} e^{\lambda_{k} t} \xi_{k}+ \sum_{k=r+1}^{r+c} 2 \operatorname{Re} [c_{k} e^{\lambda_{k} t} \xi_{k}] $$In conclusion, in equation (1) above, $c_{m,i}$, $a_{m,j}$, $b_{m,j}$ are "connected" in such a way that they only suppose $n$ degrees of freedom: they are linear functions of real and imaginary parts of $c_k$ and we have that the first $r$ $c_k$s are real and the last $2c$ are paired by conjugation .
Sometimes it is preferred other expression of the general solution. For $k=r+1..r+c$ the terms above can be rewritten:
$$ 2Re[c_{k} e^{\lambda_{k} t} \xi_{k}]= Re(\xi_k)e^{\alpha_k t} \cdot \left(2Re(c_k)cos(\omega_k t)-2Im(c_k)Sin(\omega_k t)\right)+ $$ $$ +Im(\xi_k)e^{\alpha_k t} \cdot \left(-2Im(c_k)cos(\omega_k t)-2Re(c_k)Sin(\omega_k t)\right)+ $$that can be clean be means of a change of names and trigonometric properties into:
$$ e^{\alpha_k t}A_k cos(\omega_k t+\phi_k) \cdot v_k^1+ e^{\alpha_k t}A_k sin(\omega_k t+\phi_k) \cdot v_k^2 $$where $v_k^1=Re(\xi_k)$, $v_k^2=Im(\xi_k)$, $A_k=|2c_k|$ and $\phi_k=arg(2 c_k))$
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Author of the notes: Antonio J. Pan-Collantes
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