In the paper Adapted solutions of backward stochastic differential equations by Pardoux and Peng the authors prove in Lemma 2.1 that the stochastic equation $$ x(t) + \int_t^1 f(s) ds + \int_t^1 (g(s) + y(s)) dW(s) = X, \quad t \in [0,1] $$ has a unique solution. Here $W(t)$ is a $k$-dimensional Brownian motion with natural filtration $\mathcal F_t$, $f \in M^2(0,1;\mathbb{R}^d)$ and $g \in M^2(0,1;\mathbb{R}^{d+k})$ ($M^2$ is the class of $\mathcal F_t$- progressively measurable processes indexed by $t \in [0,1]$ which are square integrable with values in $\mathbb{R}^d$ and $\mathbb{R}^{d+k}$ respectively), and $X$ is $\mathcal F_1$ measurable. By solution to the equation we mean $(x(t),y(t))$ (with values in $\mathbb{R}^d \times \mathbb{R}^{d+k}$) which is $\mathcal F_t$-adapted.
The unique solution, constructed in the proof, is given by $$ x(t) = E(X - \int_t^1 f(s) ds|\mathcal F_t), \quad t \in [0,1], $$ and $$ y(t) = \overline y(t) - g(t), \quad t \in [0,1], $$ where $\overline y(t) \in M^2(0,1;\mathbb{R}^{d+k})$ satisfies $$ \int_0^t \overline y(s) dW(s) = E(X - \int_0^1 f(s) ds| \mathcal F_t) - x(0), \quad t \in [0,1]. $$
How can I show that such $(x(t),y(t))$ is a solution? How can I show that it is unique?
Here is how I see it by the definition of $y$ you have $\forall t>0$ : $$X=x(t) + \int_t^1 f(s) ds + \int_t^1 (g(s) + y(s)) dW(s)$$ $$X=x(t) + \int_t^1 f(s) ds + \int_t^1 \bar{y}(s) dW(s) (1)$$ And the term $$\int_t^1 \bar{y}(s) dW(s)=\int_0^1 \bar{y}(s) dW(s)-\int_0^t \bar{y}(s) dW(s)$$
$$= E(X - \int_0^1 f(s) ds| \mathcal F_1) - x(0) - E(X - \int_0^1 f(s) ds| \mathcal F_t) + x(0)$$ $$= X - \int_0^1 f(s) ds +\int_0^t f(s) ds - E(X - \int_t^1 f(s) ds| \mathcal F_t) $$
becomes in the end : $$= X - \int_t^1 f(s) ds - x_t $$ (because by definition : $x_t=E(X - \int_t^1 f(s) ds| \mathcal F_t) $
Putting this expression back into (1) everything collapse which proves the claim unless mistaken.