Disclaimer: I am a total beginner in stochastic calculus
Let $\mathrm{d}X_t = a(b-X_t) \,\mathrm{d}t + c X_t \, \mathrm{d}W_t$ be a stochastic differential equation (this SDE is the variance differential form for the standard GARCH(1,1) model) where $a$, $b$, and $c$ are positive constants, so I tried to solve it but I'm sure I made a mistake in the process, here is my attempt:
$$\mathrm{d}X_t = a(b-X_t) \, \mathrm{d}t + c X_t \, \mathrm{d}W_t$$ $$\mathrm{d}X_t = ab \, \mathrm{d}t - aX_t \, \mathrm{d}t + c X_t \, \mathrm{d}W_t$$ $$\mathrm{d}X_t + aX_t \, \mathrm{d}t - c X_t \mathrm{d}W_t = ab \, \mathrm{d}t$$ $$\int_0^t \mathrm{d}X_t + \int_0^t aX_t \, \mathrm{d}t - \int_0^t c X_t \, \mathrm{d}W_t = \int_0^t ab \, \mathrm{d}t$$ $$\int_0^t \mathrm{d}X_t + \left[\int_0^t aX_t \, \mathrm{d}t + \int_0^t (-c) X_t \, \mathrm{d}W_t\right] = \int_0^t ab \, \mathrm{d}t \tag 1$$
(The following step and conclusion is where I must've definitely made the mistake)
The term inside the brackets looks 
and has the following solution:
$$X_0 \mathrm{e}^{\left[\left(a - \frac{(-c)^2}{2}\right)t - cW_t\right]} \tag 2$$
We replace (2) in (1) and we get,
$$\int_0^t \mathrm{d}X_t + X_0 \mathrm{e}^{\left[\left(a - \frac{(-c)^2}{2}\right)t - c W_t\right]} = \int_0^t ab \mathrm{d}t$$ $$X_t - X_0 + X_0 \mathrm{e}^{\left[\left(a - \frac{(-c)^2}{2}\right)t - c W_t\right]} = abt$$ $$X_t - X_0 \left(1 - \mathrm{e}^{\left[\left(a - \frac{(-c)^2}{2}\right)t - c W_t\right]}\right) = abt$$
We finally get,
$$X_t = X_0 \left(1 - \mathrm{e}^{\left[\left(a - \frac{(-c)^2}{2}\right)t - c W_t\right]}\right) + abt$$
