let us assume the following ODE
$$ \frac{dA}{dt} +gA = X $$
where $X \sim N(0,1)$ and $g$ is a factor. $X$ changes its value in every timestep $dt$ which means that $X$ is a stochastic process.
if we would simply solve this ODE deterministically, then the integrating factor would be found quickly:
$$A = \frac{\int e^{gt}Xdt+C}{e^{gt}}$$
where one knows that
$$\int e^{gt}Xdt = \int e^{gt}dB_t $$ with standard Brownian Motion $B_t$
So:$$A = \frac{\int e^{gt}dB_t+C}{e^{gt}}$$
When i compute the mean and variance of $A$ i get
$$\mathbb{E}(A_t) = Ce^{-gt}$$ $$\mathbb{V}(A_t) = \frac{1}{gt}$$
but when i interprete this ODE as an Ornstein Uhlenbeck process i get the same mean, but different variance...
where is the problem?
Your $C$ is not a constant, though it is independent of $t$; it is random variable. So you cannot find the variance by the first method.