I don't think that you have to transfer your differential equation to first order to get the frequency response (solution of @Jagerber)
Time domain
$$\mathbf M\,\mathbf{\ddot{x}}+\mathbf{K}\mathbf x+
\mathbf C\,\mathbf{\dot{x}}=\mathbf f(t)$$
transformed to Laplace domain and solve for $\mathbf{X}(s)$
$$\mathbf X(s)=\underbrace{\left[s^2\,\mathbf M+\mathbf K+s\,\mathbf C\right]^{-1}}_{\mathbf H(s)}\,\mathbf F(s)$$
$~\mathbf H(s)~$ is a matrix $n\times n$ where n is the number of the ODE's
Example:
$$m\,\ddot x_1+k\,(x_1-x_2)+c\,\dot x_1=f_1(t)\\
m\,\ddot x_2-k\,(x_1-x_2)+c\,\dot x_2=f_2(t)$$
$$\mathbf M=\left[ \begin {array}{cc} m&0\\ 0&m\end {array}
\right]\\
\mathbf K= \left[ \begin {array}{cc} k&-k\\ -k&k\end {array}
\right] \\
\mathbf C=\left[ \begin {array}{cc} c&0\\ 0&c\end {array}
\right]
$$
$\Rightarrow$
$$\mathbf H=\left[ \begin {array}{cc} {\frac {m{s}^{2}+cs+k}{s \left( {m}^{2}{s}^
{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2\,ck \right) }}&{\frac {k}{s \left(
{m}^{2}{s}^{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2\,ck \right) }}
\\ {\frac {k}{s \left( {m}^{2}{s}^{3}+2\,m{s}^{2}c+2
\,msk+{c}^{2}s+2\,ck \right) }}&{\frac {m{s}^{2}+cs+k}{s \left( {m}^{2
}{s}^{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2\,ck \right) }}\end {array}
\right]
$$
thus you obtain 4 transfer functions and
$$X_1(s)=H_{1,1}(s)F_1(s)+H_{1,2}(s)\,F_2(s)\\X_2(s)=H_{2,1}(s)F_1(s)+H_{2,2}(s)\,F_2(s)$$
edit
solution of @Jagerber
with
$$x_1=y_1~,x_2=y_2~,\dot x_1=y_3~,\dot x_2=y_4$$
\begin{align*}
&\mathbf{\dot{y}}=\mathbf{A}\,\mathbf{y}+\mathbf{b}\\
&\left[ \begin {array}{c} {\dot y}_{{1}}\\ {\dot y}_
{{2}}\\ {\dot y}_{{3}}\\ {\dot y}_
{{4}}\end {array} \right]
=\left[ \begin {array}{cccc} 0&0&1&0\\ 0&0&0&1
\\ -{\frac {k}{m}}&{\frac {k}{m}}&-{\frac {c}{m}}&0
\\ {\frac {k}{m}}&-{\frac {k}{m}}&0&-{\frac {c}{m}}
\end {array} \right]
\,\left[ \begin {array}{c} y_{{1}}\\ y_{{2}}
\\ y_{{3}}\\ y_{{4}}\end {array}
\right]
+ \left[ \begin {array}{c} 0\\ 0\\
{\frac {f_{{1}}}{m}}\\ {\frac {f_{{2}}}{m}}
\end {array} \right]\\\\
&\text{transfer to Laplace and solve for $~Y(s)$}\\\\
&\mathbf{Y}(s)=\underbrace{\left[s\,I_4-A\right]^{-1}}_{\mathbf{H}(s)}\,\mathbf{b}(s)
\end{align*}
with
the differential equations transfer to first order differential equations
\begin{align*}
&\mathbf{H}(s)=\left[ \begin {array}{cccc} {\frac {m{s}^{2}+cs+k}{ \left( cs+m{s}^{2
}+2\,k \right) s}}&{\frac {k}{ \left( cs+m{s}^{2}+2\,k \right) s}}&{
\frac {m \left( m{s}^{2}+cs+k \right) }{s \left( {m}^{2}{s}^{3}+2\,m{s
}^{2}c+2\,msk+{c}^{2}s+2\,ck \right) }}&{\frac {km}{s \left( {m}^{2}{s
}^{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2\,ck \right) }}
\\ {\frac {k}{ \left( cs+m{s}^{2}+2\,k \right) s}}&{
\frac {m{s}^{2}+cs+k}{ \left( cs+m{s}^{2}+2\,k \right) s}}&{\frac {km}
{s \left( {m}^{2}{s}^{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2\,ck \right) }}
&{\frac {m \left( m{s}^{2}+cs+k \right) }{s \left( {m}^{2}{s}^{3}+2\,m
{s}^{2}c+2\,msk+{c}^{2}s+2\,ck \right) }}\\ -{\frac
{k}{cs+m{s}^{2}+2\,k}}&{\frac {k}{cs+m{s}^{2}+2\,k}}&{\frac {m \left(
m{s}^{2}+cs+k \right) }{{m}^{2}{s}^{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2
\,ck}}&{\frac {km}{{m}^{2}{s}^{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2\,ck}}
\\ {\frac {k}{cs+m{s}^{2}+2\,k}}&-{\frac {k}{cs+m{s}
^{2}+2\,k}}&{\frac {km}{{m}^{2}{s}^{3}+2\,m{s}^{2}c+2\,msk+{c}^{2}s+2
\,ck}}&{\frac {m \left( m{s}^{2}+cs+k \right) }{{m}^{2}{s}^{3}+2\,m{s}
^{2}c+2\,msk+{c}^{2}s+2\,ck}}\end {array} \right]
\end{align*}
thus the transfer function for example $~Y_3~$
\begin{align*}
Y_3(s)=\sum_{i=1}^{4} H_{3,i}\,b_i=
{\frac { \left( m{s}^{2}+cs+k \right) F_{{1}}}{{m}^{2}{s}^{3}+2\,m{s}^
{2}c+2\,msk+{c}^{2}s+2\,ck}}+{\frac {kF_{{2}}}{{m}^{2}{s}^{3}+2\,m{s}^
{2}c+2\,msk+{c}^{2}s+2\,ck}}
\end{align*}
must be equal to the transfer function $~X_1(s)$ (my solution) this is the case so my approach is correct!!
who ever vote it down please let me know why ??