Im really new in the stochastic procceses please help me.
How can I solve this stochastic differential equation?
$$dX = A(t)Xdt$$ $$X(0) = X_0$$
If $A$:[0,$\infty$]$\to$ $R$ is continous and $X$ is a real random variable.
Im really new in the stochastic procceses please help me.
How can I solve this stochastic differential equation?
$$dX = A(t)Xdt$$ $$X(0) = X_0$$
If $A$:[0,$\infty$]$\to$ $R$ is continous and $X$ is a real random variable.
On
The (deterministic) ODE
$$\frac{dx(t)}{dt} = A(t) \cdot x(t) $$
can be solved by transforming it into
$$\log x(t)-c = \int^t \frac{1}{x(s)} dx(s) = \int^t A(s) \, ds$$
The same approach applies to this stochastic differential equation. We use Itô's formula for $f(x) := \log x$ and the Itô process $X_t$ and obtain
$$\log X_t - \log x_0 = \int_0^t \frac{1}{X_s} \, dX_s = \int_0^t A(s) \, ds$$
(Note that the second term in Itô's formula equals $0$.) Consequently,
$$X_t = X_0 \cdot \exp \left( \int_0^t A(s) \, ds \right)$$
This means that the solution of the ODE and the SDE coincide and as @user39360 already remarked that's not surprising since there is no stochastic part in your stochastic differential equation at all - and so the stochastic differential equation becomes an ordinary differential equation.
Let me finally remark that the approach described above (i.e. considering the corresponding ordinary differential equation and adapting the idea of the solution method in the determinstic case) works often quite well in a more general setting.
You would solve this equation like any other differential equation (assuming $X(t)\in\mathbb{R}^n\times[0,\infty)$),
$X'(t) = A(t)X(t) \implies X(t) = \exp(A(t))X_0.$
The randomness of $X(t)$ must enter either through the initial condition $X_0$ or the function $A(t)$.