Solving an initial value problem for the PDE $u_t + u_x = (x+t) \cos(xt)$ This is my first time to study PDE in grad level . I have to solve this IVP :
$$
u_t + u_x = (x+t) \cos(xt)  \tag{*}
$$ 
on $\mathbb{R} \times (0,\infty)$.
$$
u(x,0) = \sin(x) \tag{**}
$$ 
on $(x,t) \in \mathbb{R} \times \{t=0\}$.
In my class note, I was given a formula to use , and I obtain:
$$
u(x,t) = \sin(x-tb) + 
\int\limits_0^t \left[x-(t-\sigma)b+\sigma\right] 
\cos\left(\left[x-(t-\sigma)b\right]\sigma\right) \, d\sigma
$$
where $b$ is a constant. 
I don't know what I need to do next. Is this one correct?
 A: Why to learn that complicated formula? Will you even remember it when taking your exam? Why not using the method of characteristics, which I think it should be firstly taught when learning PDEs. In your case, it reads:
$$\frac{\mathrm{d}t}{1} = \frac{\mathrm{d}x}{1} = \frac{\mathrm{d}u}{(x+t) \cos{xt}}.  $$
From the first equatlity we have: $\mathrm{d}t - \mathrm{d}x = 0, $ which after integrating yields $x-t=c$, where $c$ is a constant called characteristic. Take now the second equality and put $x$ as a function of $t$ to have:
$$\mathrm{d}t = \frac{\mathrm{du}}{(2t+c) \cos{[(t+c)t}]},$$ 
whichi is a separable differential equation. Solve the corresponding integral$^*$ and put the constant of integration as a function of $c = x-t$, solve for the given initial condition and you are done!
Cheers!

$^*$Note that the argument of the $\sin$ is a primitive of $2t+c$.
A: Just checking if your solution candidate fullfills the PDE $(*)$ and boundary condition $(**)$. 
The condition $(**)$ holds.
For the PDE we need to calculate $u_x$ and $u_t$. For calculating $u_t$ we use the Leibniz rule.
We get
$$
\begin{align}
u_x(x,t) 
&= 
\cos(x-tb) + 
\frac{\partial}{\partial x} 
\int\limits_0^t \left[x-(t-\sigma)b+\sigma\right]
\cos\left(\left[x-(t-\sigma)b\right]\sigma\right) \, d\sigma \\
&=
\cos(x-tb) + 
\int\limits_0^t 
\frac{\partial}{\partial x} 
\left[x-(t-\sigma)b+\sigma\right] 
\cos\left(\left[x-(t-\sigma)b\right]\sigma\right) \, d\sigma \\
&=
\cos(x-tb) + \\
&\int\limits_0^t 
\cos\left(\left[x-(t-\sigma)b\right]\sigma\right) -
\left[x-(t-\sigma)b+\sigma\right]
\sin\left(\left[x-(t-\sigma)b\right]\sigma\right) \sigma 
\, d\sigma
\end{align}
$$
and
$$
\begin{align}
u_t(x,t) 
&= 
-b \cos(x-tb) + 
\frac{\partial}{\partial t} 
\int\limits_0^t \left[x-(t-\sigma)b+\sigma\right]
\cos\left(\left[x-(t-\sigma)b\right]\sigma\right) \, d\sigma \\
&=
-b \cos(x-tb) + 
\int\limits_0^t 
\frac{\partial}{\partial t} 
\left[x-(t-\sigma)b+\sigma\right] 
\cos\left(\left[x-(t-\sigma)b\right]\sigma\right) \, d\sigma + \\
& \left[x-(t-t)b+t\right] 
\cos\left(\left[x-(t-t)b\right]t\right) \frac{dt}{dt} \\
&=
-b\cos(x-tb) + \\
&(-b) \int\limits_0^t 
\cos\left(\left[x-(t-\sigma)b\right]\sigma\right) -
\left[x-(t-\sigma)b+\sigma\right]
\sin\left(\left[x-(t-\sigma)b\right]\sigma\right) \sigma \, d\sigma + \\
& (x+t) \cos(xt)
 \\
&=
-b \, u_x(x,t) + (x+t) \cos(xt)
\end{align}
$$
So, assuming I made no mistake, your candidate solves
$$
u_t(x,t) + b \, u_x(x,t) = (x+t) \cos(xt)
$$
which is your $(*)$ if $b = 1$.
A: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dx}{ds}=1$ , letting $x(0)=x_0$ , we have $x=s+x_0=t+x_0$
$\dfrac{du}{ds}=(x+t)\cos xt=(2s+x_0)\cos((s+x_0)s)$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)+\int_0^s(2\sigma+x_0)\cos((\sigma+x_0)\sigma)~d\sigma=f(x-t)+\int_0^t(2\sigma+x-t)\cos((\sigma+x-t)\sigma)~d\sigma$
$u(x,0)=\sin x$ :
$f(x)=\sin x$
$\therefore u(x,t)=\sin(x-t)+\int_0^t(2\sigma+x-t)\cos((\sigma+x-t)\sigma)~d\sigma$
