Solving the SDE $dX(t) = (c(t) + d(t)X(t))dt + (e(t) + f(t)X(t))dW(t)$ How to solve 
$dX(t) = (c(t) + d(t)X(t))dt + (e(t) + f(t)X(t))dW(t)$ together with the initial condition $X(0) = X_0$.
 A: $$dX_t = (c(t)+d(t) X_t) \, dt + (e(t)+f(t) X_t) \, dW_t, \quad X_0=x_0 \tag{1}$$
Hints


*

*Solve the homogeneous linear SDE $$dY_t = d(t) Y_t \, dt + f(t) Y_t \, dW_t, \quad Y_0 = y_0 \tag{2}$$ In order to do so, apply Itô's formula to $Z_t := \ln Y_t$. 

*Let $X_t^0$ such that $\frac{1}{X_t^0}$ solves $(2)$ 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 $X_t = \frac{Z_t}{X_t^0}$.

A: Here is a rough version of what I've done.
I remain open to any remark.


*

*We start by solving the homogeneous associated SDE:
\begin{equation}
  \left\{
 \begin{aligned}
  dY_t &= d(t) Y_t \, dt + f(t) Y_t \, dW_t\\
 Y_0 &= y_0
      \end{aligned}
    \right.
\end{equation}
In order to do so, we apply Ito's formula to $Z_t := \ln Y_t$.
\begin{align*}
  dZ_t = [d(t)- \tfrac 12 f^2(t)]dt + f(t)dW_t
 \end{align*}
Integrating we get
\begin{align*}
  ln(\frac{Y_t}{Y_0})=Z_t-Z_0= \int_0^t \! \left(d(s)- \tfrac 12 f^2(s)\right)\mathrm{d}s + \int_0^t \! f(s)\mathrm{d}W_s
 \end{align*}
and hence
\begin{align*}
  Y_t= y_0 \exp \left( \int_0^t \! \left(d(s)- \tfrac 12 f^2(s)\right)\mathrm{d}s + \int_0^t \! f(s)\mathrm{d}W_s \right)
 \end{align*}

*Now let $U_t$ such that $U_t^{-1}$ solves $(3.2)$ with initial condition $U_0=1$. 
Then we have $$ U_t=\exp \left( -\int_0^t \! \left(d(s)- \tfrac 12 f^2(s)\right)\mathrm{d}s - \int_0^t \! f(s)\mathrm{d}W_s \right)$$
Define $$I_t =\int_0^t \! f(s)\mathrm{d}W $$ or equivalently $$dI_t = f(t)dW_t, \ \, I_0=0 \ \ \  \ \text{  (so } \mu = 0, \sigma = f \text{  )}$$
Then we have $$U_t=g(t,I_t)$$ where $$g(t,x)= \exp \left( -\int_0^t \! \left(d(s)- \tfrac 12 f^2(s)\right)\mathrm{d}s -x \right)$$
By Ito's lemma one has
\begin{align*}
 dU_t = dg(t,I_t)&= U_t \left ( \frac{\partial g}{\partial t} + \mu \frac{\partial g}{\partial x}+ \tfrac 12 \sigma^2 \frac{\partial^2 g}{\partial x^2}\right)dt + \sigma \frac{\partial g}{\partial x} dW_t\\
 &= U_t \left [  \left(f^2(t)-d(t)\right) dt - f(t)dW_t\right]
\end{align*}
Finally, let $Z_t=X_tU_t$
Again Ito's formula (more precisely the product rule) gives
    \begin{align*}
dZ_t  &= dX_tU_t + dU_tX_t +d<X,U>_t\\
   &= \left[\left(c(t)+ \cancel{d(t)X_t} \right)dt+\left(e(t)+ \cancel{f(t)X_t} \right)dW_t\right] U_t\\
   &+ \left[\left(\cancel{f^2(t)} -\cancel{d(t)} \right)dt+\  - \cancel{f(t) dW_t}\right] X_tU_t\\
   &+ \left(e(t) + \cancel{f(t)X_t} \right) \left(-f(t)U_t\right) dt\\
   &= \left(c(t) - e(t)f(t) \right)U_tdt + e(t)U_tdW_t
\end{align*}
Integrating results in
    \begin{align*}
  Z_t-Z_0 &=\int_0^t \! \left(c(s) - e(s)f(s) \right)U_s\mathrm{d}s + \int_0^t \!e(s)U_s\mathrm{d}W_s\\
  X_tU_t-X_0U_0 &=\int_0^t \! \left(c(s) - e(s)f(s) \right)U_s\mathrm{d}s + \int_0^t \!e(s)U_s\mathrm{d}W_s\\
   \end{align*}
            and hence 
\begin{align*}
  X_t &= U_t^{-1} \left[ x_0 + \int_0^t \! \left(c(s) - e(s)f(s) \right)U_s\mathrm{d}s + \int_0^t \!e(s)U_s\mathrm{d}W_s  \right]
 \end{align*}
is the solution for $(3.1)$ where
$$ U_t^{-1}= \frac{1}{U_t}=\exp \left( -\int_0^t \! \left(d(s)- \tfrac 12 f^2(s)\right)\mathrm{d}s - \int_0^t \! f(s)\mathrm{d}W_s \right)$$
