The following is an old exam question I think might be misstated. Consider the SDE
$$dX(u)=(a(u)+b(u)X(u))\,du+(\gamma(u)+\sigma(u)X(u))\,dW(u)$$
where $W(u)$ is a brownian motion relative to filtration $\mathcal{F}(u)$, and $a(u), \; b(u), \; \gamma(u),$ and $\sigma(u)$ are processes adapted to this filtration. Fix an initial time $t\geq0$ and an initial position $x \in \mathbb{R}$. Define
$$Z(u)=\exp\left(\int_t^u \sigma(v)\,dW(v)+\int_t^u(b(v)-\frac{1}{2}\sigma^2(v))\,dv\right)$$
$$Y(u)=x+\int_t^u\frac{a(v)-\sigma(v)\gamma(v)}{Z(v)}\,dv+\int_t^u\frac{\gamma(v)}{Z(v)}\,dW(v)$$
Show that $X(u)=Y(u)Z(u)$ satisfies the stochastic differential equation such that $X(t)=x$. It is easy to see that the initial condition is satisfied due the bounds on the integrals. However when I differentiate $X(u)$ using Ito's product rule I don't get the SDE above, I get the SDE and an extra term $-\frac{1}{2}\sigma^2(u) \, du$. Is the problem misstated or am I in error. I believe it is misstated, here is why. To check to see if $X(u)$ solves the given SDE we differentiate and see if we come up with the SDE above. Apply Ito's product rule $$dX(u)=d(Y(u)Z(u))=Y(u)dZ(u)+Z(u)dY(u)+dY(u)dZ(u)$$ For the desired derivatives we have $$dY(u)=\frac{a(u)-\sigma(u)\gamma(u)}{Z(u)}du+\frac{\gamma(u)}{Z(u)}dW(u)$$ $$dZ(u)=Z(u)\left(\sigma(u)dW(u)+\left(b(u)-\frac{1}{2}\sigma^2(u)\right)du\right)$$ We now plug into the product rule and use the fact that $du\;du=du\;dW(u)=0$ and $dW(u)\;dW(u)=du$ to get $$dX(u)=(a(u)-\frac{1}{2}\sigma^2(u)+b(u)X(u))\,du+(\gamma(u)+\sigma(u)X(u))\,dW(u)$$ which leaves me to wonder if the term $-\frac{1}{2}\sigma^2(u)$ was left out of the original statement by accident. So is there an error in the original statement of the problem or am I doing something wrong?