Method of characteristics - heat convection I have following PDE:
\begin{align}
v \cdot \frac{\partial T}{\partial x} + \frac{\partial T}{\partial t} &= k \cdot (T-T_0)\\
T(x,0) &= T_0\\
T(0,t) &= T_1
\end{align}
It's first order PDE, so I tried to solve it by the method of characteristics:
$$
\frac{dx}{v} = \frac{dt}{1} = \frac{dT}{k(T-T_0)}
$$
From $\frac{dx}{v} = \frac{dt}{1}$ I deduced:
\begin{align}
x &= vt + C\\
C_1 &= x - vt
\end{align}
From $\frac{dt}{1}=\frac{dT}{k(T-T_0)}$ I deduced:
\begin{align}
T-T_0 &= C_2 \cdot e^{k\cdot t}\\
C_2 &= \frac{T-T_0}{e^{kt}}
\end{align}

The problem
Now I have $C_1$ and $C_2$ and I'm not sure how I'm supposed to apply both conditions.
My attempt looks like this:
I've tried to express $t$ with $C_1$:
$$
t = \frac{x-C_1}{v}
$$
And substitute it into the second expression with $C_2$:
$$
C_2 = \frac{T-T_0}{e^{k \frac{x-c_1}{v}}}
$$
But when I try to apply the first condition:
$$
C_2 = \frac{T_0-T_0}{e^{0}} = \frac{0}{1} = 0
$$
And then the second one:
$$
0 = \frac{T_1-T_0}{e^{\frac{-kC_1}{v}}}
$$

I know, that in the method of characteristics I'm supposed to parametrize the T(x,t) function somehow, but I don't know, how should I do this when the equation is not homogeneous? Can you, please, help me?
 A: Taking it from your results :
$$C_1 = x - vt$$
$$C_2 = \frac{T-T_0}{e^{kt}}$$
The general solution on implicit form is :
$$\Phi\left(x-vt\:,\:\frac{T-T_0}{e^{kt}}\right)=0$$
where $\Phi(X,Y)$ is any differentiable function of two variables.
Solving $\Phi(X,Y)=0$ for $Y$ leads to the general solution on the explicit form :
$$\frac{T-T_0}{e^{kt}}=F(x-vt)$$
$F$ is any differentiable function.
$$T(x,t)=T_0+e^{kt}F(x-vt)$$
Up to this point, this is consistent with your results. 
The hitch comes from the conditions : $\begin{cases} T(x,0)=T_0 \\T(0,t)=T_1 \end{cases}$
We observe that $T(0,0)=T_0=T_1$ . This is contradictory if $T_0\neq T_1$
As a consequence they are two cases : 
First case : $T_0=T_1\quad \to\quad F(x)=0 \quad\to\quad$ The solution is trivial : $T(x,t)=T_0$
Second case : $T_0\neq T_1 \quad\to\quad$The particular point $(x=0\:,\:y=0)$ must be excluded from the conditions. 
The conditions become consistent thanks to the Heaviside step function $H$ :
$$T(0,t)=T_0+(T_1-T_0)H(t)$$
$T_0+(T_1-T_0)H(t)=T_0+e^{kt}F(0-vt) \quad\to\quad F(-vt)=(T_1-T_0)H(t)e^{-kt}$
Change of variable : $X=-vt\quad\to\quad t=-\frac{X}{v}$
$$F(X)=(T_1-T_0)H\left(-\frac{X}{v}\right)e^{\frac{k}{v}X}$$
$$F(x-vt)=(T_1-T_0)H\left(-\frac{x-vt}{v}\right)e^{\frac{k}{v}(x-vt)}$$
$$T(x,t)=T_0+e^{kt}(T_1-T_0)H\left(-\frac{x-vt}{v}\right)e^{\frac{k}{v}(x-vt)}$$
$$T(x,t)=T_0+(T_1-T_0)H\left(t-\frac{x}{v}\right)e^{\frac{k}{v}x}$$
Check of conditions :
When $t=0$ and $x>0$ then $H\left(-\frac{x}{v}\right)=0\quad\to\quad T(x,0)=T_0$
When $x=0$ and $t>0$ then $H\left(t\right)e^{\frac{k}{v}0}=1\quad\to\quad T(0,t)=T_0+(T_1-T_0)=T_1$ 
