Showing that $X_t=\int_0^t e^{A(t-s)}KX_s\,dB_s+\int_0^te^{A(t-s)}M\,dB_s$ satisfies $dX_t$ Consider $$dX_t=AX_t\,dt+KX_t\,dB_t+M\,dB_t,$$ where $A=\begin{pmatrix} 0 & 1 \\ -w^2 & -a_0\end{pmatrix}$, $K=a_o\eta\begin{pmatrix} 0 & 0 \\ 0 & -1\end{pmatrix}$ and $M=T_0\eta\begin{pmatrix} 0 \\ 1\end{pmatrix}$ for constants $a_0,w,T_0,a_0$ and $\eta$. Here $B_t$ denotes a Brownian motion at time $t$ ($B_0=0$).
I am trying to show that $X_t$ satisfies the integral equation
$$X_t=\int_0^t e^{A(t-s)}KX_s\,dB_s+\int_0^te^{A(t-s)}M\,dB_s,$$ if $X_0=0$.
Clearly $X_t$ satisfies the initial condition. To see if $X_t$ satisfies $dX_t$, I attempt to write $X_t$ in the form of $dX_t$.
\begin{align}
X_t&=X^1_t+X^2_t  \\
\implies dX_t&=dX^1_t+dX^2_t  \\
  &=\left(KX_t\,dB_t+AX^1_t\right)+\left(M\,dB_t+AX^2_t\right) \\
  &=AX_t+KX_t\,dB_t+M\,dB_t.
\end{align}
For some reason the first term in the final line is missing a $dt$. How does this appear?
This question is taken from Oksendal's Stochastic Differential Equations Vol. $5$, question $5.13$
Update
With thank to TheBridge, I have tried to use the multidimensional Ito Lemma. Let $X_t=(X^1_t,X^2_t)$. Then
\begin{align}
X_t&=\int_0^t e^{A(t-s)}KX_s\,dB_s+\int_0^te^{A(t-s)}M\,dB_s \\
e^{-At} X_t&=\int_0^t e^{-As}KX_s\,dB_s+\int_0^te^{-As}M\,dB_s \\
e^{-At}X^1_t+e^{-At}X^2_t&=\int_0^t e^{-As}KX_s\,dB_s+\int_0^te^{-As}M\,dB_s
\end{align}
I let $g(t,x_1,x_2)=e^{-At}x_1+e^{-At}x_2$. Then by Ito's multidimensional lemma,
\begin{align}
d\left(e^{-At}X_t\right)&=\frac{\partial g}{\partial t} (t,X^1_t,X^2_t)\,dt + 
\left(\frac{\partial g}{\partial x_1}(t,X^1_t,X^2_t)\,dX^1_t+\frac{\partial g}{\partial x_2}(t,X^1_t,X^2_t)\,dX^2_t\right)+\frac{1}{2}\left(\frac{\partial^2 g}{\partial x_1^2}(t,X^1_t,X^2_t)\,\left(dX^1_t\right)^2+\frac{\partial^2 g}{\partial x_2^2}(t,X^1_t,X^2_t)\,\left(dX^2_t\right)^2\right) \\
&=-Ae^{-At}X(t)\,dt+e^{-At}\,dX_1(t)+e^{-At}\,dX_2(t)
\end{align}
 A: Well
Let's write $g(t,x_1,x_2) = e^{-A.t}(x_1,x_2)$ then by multidimensional Itô's lemma :
$$d\left(e^{-At}X_t\right)=$$(1)
$$\frac{\partial g}{\partial t} (t,X^1_t,X^2_t)\,dt  $$ (2)
$$+\left( \nabla g(t,X^1_t,X^2_t)\,dX_t\right)$$(3)
$$+\frac{1}{2}\left(dX_t\right)^T\Delta g(t,X^1_t,X^2_t)\,\left(dX_t\right) $$
Where $\nabla g$ is the gradient of $g$ with respect to vector $X$ and $\Delta$ is the Laplacian.
So $(1) =-Ae^{-At}X(t)\,dt$
$(3)= (0,0)$ because the Laplacian with respect to $(x=(x_1,x_2)$ of the function $(x_1,x_2)$ is the $0$ 2 by 2 matrix.
We are left with :
$(2)=e^{-At}.Id.dX_t$ where Id is the 2 dim identity matrix.
And putting it all together we have :$$d\left(e^{-At}X_t\right)= -Ae^{-At}X(t)\,dt +e^{-At}.dX_t$$ $$=-Ae^{-At}X(t)\,dt+ e^{-At}.(AX_t\,dt+KX_t\,dB_t+M\,dB_t)$$
$$=e^{-At}.(KX_t\,dB_t+M\,dB_t)$$
Now integrate (knowing $X_0=0$ at hypothesis in Oksendal's exercise you forgot to mention) :
$$e^{-At}X_t-1.X_0=e^{-At}X_t=\int_0^t e^{-As}KX_s\,dB_s+\int_0^t e^{-As}M\,dB_s)$$
So $$X_t=\int_0^t e^{A(t-s)}KX_s\,dB_s+\int_0^t e^{A(t-s)}M\,dB_s)$$
N.B. : $e^{A}$ needs to be invertible.
QED
