Let $u(t, x) = E_x[\int_0^t \! 1_{[-1,1]}(B(s))ds] = E[$Time spent by B(s) in $[-1, 1]$ up to time $t$ | $B(0) = x$]. write a differential equation for $u(t,x).$ Include appropriate boundary conditions.
Is this an application of Feynman-Kac formula? I think I was told to use an the identity $E(X)=\frac{\partial}{\partial \lambda}E(e^{\lambda x})$ evaluated at $\lambda = 0$. I don't know how to go from here.
Define $$v(t,x,y)=E_x[e^{\int_0^ty\mathbf1_{[-1,1]}(B(s))\ \mathrm{d}s}]$$ A simple application of Feynman-Kac shows that $v$ solves the PDE $$v_t=\tfrac12v_{xx}+y\mathbf1_{[-1,1]}(x)v,\\ v(0,x,y)=1,\\ v(t,x,0)=1.$$ Differentiating the first two lines with respect to $y$ (and assuming appropriate continuity of the derivatives), $$(v_y)_t=\tfrac12(v_y)_{xx}+(yv_y+v)\mathbf1_{[-1,1]}(x),\\ v_y(0,x,y)=0.$$ Now by your hint we know that $u(t,x)=v_y(t,x,0)$. Evaluating everything at $y=0$, it follows that $u$ solves the PDE $$u_t=\tfrac12u_{xx}+\mathbf1_{[-1,1]}(x),\\ u(0,x)=0.$$