Solving the SDE $dX(t) = (c(t) + d(t)X(t))dt + (e(t) + f(t)X(t))dW(t)$

$$dX_t = (c(t)+d(t) X_t) \, dt + (e(t)+f(t) X_t) \, dW_t, \quad X_0=x_0 \tag{1}$$

Hints

1. 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 $Z_t := \ln Y_t$.
2. 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 $X_t = \frac{Z_t}{X_t^0}$.

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

1. 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*}

2. 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 $$ U_t^{-1}= \frac{1}{U_t}=\exp \left( -\int_0^t \! \left(d(s)- \tfrac 12 f^2(s)\right)\mathrm{d}s - \int_0^t \! f(s)\mathrm{d}W_s \right)$$