Solve the linear SDE $dX_t = aX_t \, dt +(b+cX_t) \, dW_t$ I am trying to find the solution to the SDE:
$$
dX_t=aX_tdt+(b+cX_t)dW_t
$$
for $t\ge0$, $X_0>0$, constants $a,b,c$
Would appreciate any hints as to how to approach this using ito's formula, I'm having trouble getting started -- choosing a function to apply Ito's to. 
 A: *

*Solve the homogeneous SDE $$dX_t = aX_t \, dt + cX_t \, dW_t. \tag{1}$$ To this end, apply Itô's formula to $Z_t := \log X_t$. 
Motivation: Dividing $(1)$ (formally) by $X_t$ yields $$\frac{dX_t}{X_t} = a \, dt + c \, dW_t.$$ If this equation was an ODE (ordinary differential equation), then we would integrate both sides (i.e. use a separation of variables) and obtain an implicit equation of the form $$\log X_t-\log X_0 = f(t,W_t).$$

*Let $(X_t^0)_{t \geq 0}$ be such that $\frac{1}{X_t^0}$ solves the homogeneous SDE $(1)$ (with initial value $X_0^0=1$). Use Itô's formula to derive the stochastic differential of $$Z_t := X_t \cdot X_t^0. \tag{2}$$ Then, $$X_t = Z_t \cdot (X_t^0)^{-1}.$$ 
Motivation: Again, if we consider an homogeneous linear ODE, then we obtain a solution of the inhomogeneous equation by the method of variation of parameters, i.e. we set $$x(t) := c(t) x_h(t)$$ where $x_h$ is the solution to the homogeneous ODE. Hence, $$c(t) = \frac{x(t)}{x_h(t)}.$$ If we compare this with $(2)$, then we see that $Z_t$ is the counterpart of $c$.
