Solve the PDE $ u_{t}+\langle(1,-1),Du(x,y,t)\rangle=\sin(t) $ Find the solution of the following equation $$u_{t}(x,y,t)+\langle(1,-1),Du(x,y,t)\rangle=\sin(t)$$ Where $(x,y,t)\in\mathbb{R}^{2}\times (0,+\infty)$ and initial condition $u(x,y,0)=1$.
I try to solve this by characteristic method, note that $u_{t}(x,y,t)+\langle(1,-1),Du(x,y,t)\rangle=\sin(t)$ is equivalent to $$u_{t}(x,y,t)+u_{x}(x,y,t)-u_{y}(x,y,t)=\sin(t)$$
So, consider the function $F(x,y,t,u_t,u_x,u_y)=u_{t}(x,y,t)+u_{x}(x,y,t)-u_{y}(x,y,t)-\sin(t)=0$, but now what?, how can apply the method?  
 A: In a strict sense, $\text Du$ is a vector with $3$ components, $(u_x, u_y, u_t)$. However, it's common for some authors to use $\text Du := \text D_{\mathbf{x}}u$, where $\mathbf{x} = (x,y)$. Another way to look at it is that $t$ is a parameter, so $u(x,y,t) = u(\mathbf{x};t)$ and $\text Du = (u_x,u_y)$. Then, the chain rule says 
$$ \frac{\text du}{\text dt} = u_t + \langle \frac{\text d\mathbf x}{\text dt},\text Du \rangle$$
Comparing this with your equation gives $\frac{\text dx}{\text dt} = 1$ and $\frac{\text dy}{\text dt} = -1$, which should get you going on method of characteristics.

To actually employ the method of characteristics, we'll look back at our equations. The chain rule gives us a hint that we should write $\mathbf x = \mathbf x(t)$ as a family of curves or trajectories. The relationship between $\mathbf x$ and $t$ is given by $\frac{\text d\mathbf x}{\text dt} = (1,-1) = \mathbf c$, which has the solution $\mathbf x = \mathbf ct + \mathbf k$, where $\mathbf k = (k_1,k_2)$ is a pair of arbitrary constants. This solution gives us a family of curves parameterized by $\mathbf k$, so basically knowing $\mathbf k$ tells you which curve you are on. Because of this, it is beneficial to solve this equation for $\mathbf k$, i.e. $\mathbf k = \mathbf x - \mathbf ct$.
The chain rule tells us that, when $\mathbf x$ is parameterized by $t$ in this way, $\frac{\text du}{\text dt} = \sin(t)$. This means that, along a given curve, $u$ evolves as $u = a-\cos(t)$ for some arbitrary constant $a$, but $a$ is not necessarily the same for all curves. Thus, we write $a = a(\mathbf k) = a(\mathbf x - \mathbf ct)$. This gives our solution as
$$ u(\mathbf x;t) = a(\mathbf x-\mathbf ct) - \cos(t)$$
We can rewrite this without vector notation as $$ u(x,y,t) = a(x-t,y+t)-\cos(t) $$
Hopefully the way I've written it gives you an idea of how to solve one of these problems in general, for any finite dimensional $\mathbf x$ and any choice of constant $\mathbf c$.
