Linear non-homogenous SDE

Question

I'm struggling to understand how to resolve the following SDE: $$dX(t)=(\sin(t)-2X(t)) dt + (1+X(t))dB(t)$$ I understand that I should use the Ito formula but I have no idea how the $F(X(t),t)$ should look like. I would be very grateful for any hints.

Answer 1

Hints

1. Solve the homogeneous linear SDE $$dY_t = -2 Y_t \, dt + Y_t \, dB_t, \quad Y_0 = y_0 \tag{1}$$ To do so, use Itô's formula for $Z_t := \ln Y_t$.
2. Let $X_t^0$ such that $\frac{1}{X_t^0}$ solves $(1)$ with initial condition $X_0^0=1$. Calculate $dX_t^0$ using step 1. Now apply Itô's formula to $Z_t := X_t \cdot X_t^0$. This gives you an integral expression for $X_t = \frac{Z_t}{X_t^0}$.

Remark: This approach works, more generally, for SDEs of the form

$$dX_t = (\alpha(t)+\beta(t) X_t) \, dt + (\gamma(t) + \delta(t) X_t) \, dB_t$$

where $\alpha,\beta,\gamma,\delta$ are (non-random) functions; see e.g.

Brownian Motion - An Introduction to Stochastic Processes - René Schilling & Lothar Partzsch; Section 19.3 (2nd edition).

Answer 2

Thank you for the clarification. From step 1 I've got: $$Y(t)=\exp(-\frac{5}{2}t + B(t)) \ \ \ \ , \ \ \ \ Y(0)=1$$ so per my understanding we can assume that $$X(t)=Y(t)V(t)$$ where $$dY(t)=\beta(t)Y(t) dt + \delta(t)Y(t) dB(t)$$ and $$dV(t)=a(t) dt + b(t) dB(t) \ \ \ \ \ \ \ , \ \ \ \ \ V(0)=X(0)=0$$ now: $$dX(t)=Y(t)dV(t) + V(t) dY(t) + dV(t) dY(t)$$ after transformations we can get $$b(t)=\frac{\gamma (t)}{Y(t)} \ \ \ \ \ \ , \ \ \ \ \ \ a(t)=\frac{\alpha (t) - \delta (t) \gamma (t)}{Y(t)}$$ so finally: $$V(t)=\int_0^t (\sin(s) - 1)(\exp(\frac{5}{2}s - B(s))) ds + \int_0^t \exp(\frac{5}{2}s - B(s)) dB(s)$$ and $$X(t)=\exp(-\frac{5}{2}t + B(t))[ \int_0^t (\sin(s) - 1)(\exp(\frac{5}{2}s - B(s))) ds + \int_0^t \exp(\frac{5}{2}s - B(s)) dB(s)]$$Am I thinking correctly? And if so, can I somehow calculate the integrals from the final solution? I want to test convergence of Euler-Maruyama Method by simulation but I'm not sure if it's possible from that form.

Edit: although I can treat these integrals as respective sums in the simulation so it shouldn't be a problem.