# Solving the SDE $dX_t=bdt+cX_t dW_t$

I want to solve the SDE $dX_t=bdt+cX_t dW_t$, $X_0=0$ for $b,c\in\mathbb R$. I start by rewriting this as

$$dX_t=(\mu_1+\mu_2 X_t )dt+(\sigma_1+\sigma_2 X_t )dW_t$$

where $\mu_1=b, \mu_2=0, \sigma_1=0, \sigma_2=c$. And the general solution for linear SDE is known to be$X_t=Y_t Z_t$ where

$$dY_t=\mu_2 Y_t dt+\sigma_2 Y_t dW_t, \quad Y_0=1$$ $$dZ_t = \frac{\mu_1-\sigma_1 \sigma_2}{Y_t } dt + \frac{\sigma_1}{Y_t} dW_t, \quad Z_0=X_0=0$$

So $$dY_t=cY_t dW_t \quad \Rightarrow \quad Y_t=\exp\left(-\frac{c^2}{2}t+cW_t\right)$$ and

$$dZ_t=\frac{b}{Y_t}dt = b \exp\left(\frac{c^2}{2}t - cW_t\right)dt$$

But now I am stuck as I am not sure how to find $Z_t$.

## 1 Answer

Actually, there is nothing left to do. From

$$dZ_t = b \exp \left( \frac{c^2}{2} t - c W_t \right) \, dt$$

it follows that

$$Z_t = \underbrace{Z_0}_{0} + b \int_0^t \exp \left( \frac{c^2}{2} s - c W_s \right) \, ds.$$

Hence,

$$X_t = Y_t \cdot Z_t = b \exp \left(- \frac{c^2}{2} t+c W_t \right) \cdot \int_0^t \exp \left( \frac{c^2}{2} s - c W_s \right) \, ds.$$