Sketching base characteristics of first order pde $$xu_{x} - uu_{t} = t$$ 
with intial data: $u(1,t) = t, −\infty < t < \infty$
So I want to sketch the base characteristics and I'm not too sure what I'm doing.
characteristic equations:
$\frac{\partial{x}}{\partial{\tau}}=x$,
$\frac{\partial{t}}{\partial{\tau}}=-u$,
$\frac{\partial{u}}{\partial{\tau}}=t$
initial conditions:
$x(0)=1$, $t(0)=\xi$, $u(0)=\xi$
$x=e^\tau$ so $\tau=lnx$
Notice that:
$\frac{\partial^2{u}}{\partial{\tau^2}}=\frac{\partial{t}}{\partial{\tau}}=-u$,
This gives me the second order ODE
$\frac{\partial^2{u}}{\partial{\tau^2}}+u=0$
Solving I get:
\begin{align}  
u&=c_{1}\cos(\tau) + c_{2}\sin(\tau)\\
t&=c_{1}\sin(\tau) - c_{2}\cos(\tau)
\end{align}
From here I don't know what I'm doing?
 A: You correctly got
$$ u_{\tau\tau} = -u $$
The initial conditions are
$$ u(0) = \xi, \quad u_\tau(0) =t(0)= \xi $$
which gives
$$ u = \xi(\cos(\tau) + \sin(\tau)) $$
and
$$ t = \xi(\cos(\tau) - \sin(\tau)) $$
Cancel out $\xi$ to get
$$ u = \frac{\cos(\tau)+\sin(\tau)}{\cos(\tau)-\sin(\tau)}t = \frac{\cos(\ln x)+\sin(\ln x)}{\cos(\ln x)-\sin(\ln x)}t $$
A: $$xu_{x} - uu_{t} = t$$
$$\frac{dx}{x}=\frac{dt}{-u}=\frac{du}{t}=d\tau$$
A first characteristic equation comes from $\frac{dt}{-u}=\frac{du}{t}$ :
$$u^2+t^2=c_1$$
$u=\pm\sqrt{c_1-t^2}$
A second characteristic equation comes from $\frac{dx}{x}=\frac{dt}{-u}=\frac{dt}{\mp\sqrt{c_1-t^2}}$ :
$\ln|x|\pm\sin^{-1}\left(\frac{t}{\sqrt{c_1}}\right)=c_2$
$$\ln|x|\pm\sin^{-1}\left(\frac{t}{\sqrt{u^2+t^2}}\right)=c_2$$
The general solution of the ODE on the form of implicit equation $c_2=F(c_1)$ is :
$$\ln|x|\pm\sin^{-1}\left(\frac{t}{\sqrt{u^2+t^2}}\right)=F(u^2+t^2)$$
where $F$ is an arbitrary function, to be determined according to the specified condition.
CONDITION : $u(1,t)=t$ .
$\ln|1|\pm\sin^{-1}\left(\frac{t}{\sqrt{t^2+t^2}}\right)=F(t^2+t^2)$
$\pm\sin^{-1}\left(\frac{t}{\sqrt{2t^2}}\right)=F(2t^2) = \pm\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)=\pm\frac{\pi}{4}$
Thus $F$ is a constant function. We put it into the above general solution :
$$\ln|x|\pm\sin^{-1}\left(\frac{t}{\sqrt{u^2+t^2}}\right)=\pm\frac{\pi}{4}$$
Solving for u :
$\frac{t}{\sqrt{u^2+t^2}}=\pm\sin\left(\pm\frac{\pi}{4}-\ln|x|\right)$
Note that squaring introduces extra solutions which will be rejected latter.
$u^2= \frac{t^2}{\sin^2\left(-\ln|x|\pm\frac{\pi}{4}\right)}-t^2= t^2\cot^2\left(-\ln|x|\pm\frac{\pi}{4}\right)$
$u=\pm t\cot\left(-\ln|x|\pm\frac{\pi}{4}\right)$
One have to check in putting into the PDE and the condition, which determines the signs both $+$.
The solution of the PDE according to the specified condition is :
$$u(x,t)= t\cot\left(-\ln|x|+\frac{\pi}{4}\right)$$
