Is there a closed form solution to the matrix differential equation $\dot X=AX+XA^T - K$? Is there a closed form solution the matrix differential equation
$\dot X = AX + XA^T - K $
for example when
$A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$
and
$K = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$
In general $A$ is square and $K$ is positive semi-definite.
As far as I understand the equation:
$\dot X = AX + XA^T$
has the closed form solution
$X(t) = e^{At}Ce^{tA^T}$
Can this be extended to $\dot X = AX + XA^T - K $?
These are similar questions, but do not answer the closed form solution question:
On the solution of one matrix differential equation
Solution of differential lyapunov equation
 A: This is a linear non-homogeneous differential equation, so the standard techniques apply.  The catch is that since it is not presented in the usual form $\dot x = Ax + b$, we would either have to put this equation into that form or do things in a coordinate-free fashion.
We can reframe this problem in the standard presentation using the vectorization operator.  In particular: if we define $x = \operatorname{vec}(X), k = \operatorname{vec}(K) \in \Bbb R^{4 \times 1}$ and vectorize both sides of the differential equation, then your differential equation becomes
$$
\dot x = (I \otimes A + A \otimes I)x - k \implies\\
\dot x = 
\pmatrix{0&1&1&0\\0&0&0&1\\0&0&0&1\\0&0&0&0} x - \pmatrix{0\\0\\0\\1}
$$
This differential equation has a closed form solution and can be solved using standard methods.  To get the answer to the original equation, it suffices to "unvectorize" our general solution $x(t)$ to the above differential equation.

Flawed earlier attempt:
We can rewrite your differential equation as
$$
\dot X -AX - XA^T = -K
$$
This is a linear differential equation since the map $X \mapsto \dot X - AX - XA^T$ is linear.  Your solution $X_h(t) = Ce^{At}e^{tA^T}$ solves the homogeneous problem, $\dot X - AX - XA^T = 0$.  It now suffices to find a particular solution to the equation 
$$
\dot X -AX - XA^T = -K
$$
Taking $X = t^2P + tQ + R$ as our Ansatz, we find that
$$
\dot X - AX - XA^T = -K \implies\\
2tP + Q - A(t^2P + tQ + R) - (t^2P + tQ + R)A^T = -K \implies\\
-t^2[AP + PA^T] + t[2P - AQ - QA^T] + [Q + K - AR - RA^T]
$$
In other words, it suffices to find matrices $P,Q,R$ satisfying the system of equations
$$
\begin{cases}
AP + PA^T = 0\\
AQ + QA^T = 2P\\
AR + RA^T = Q + K
\end{cases}.
$$
For such $P,Q,R$, we can write our general solution as $X(t) = Ce^{At}e^{tA^T} + t^2P + tQ + R$.
