How to solve this partial differential equation $\frac{\partial p(k,t)}{\partial t}+k\frac{\partial p(k,t)}{\partial k}+k^2p(k,t)=0$ How to solve this partial differential equation
$$\frac{\partial p(k,t)}{\partial t}+k\frac{\partial p(k,t)}{\partial k}+k^2p(k,t)=0$$
I'm a beginner to PDE, I think I need to construct the characteristic equation but I have no clue.
 A: This method poses an auxiliary system of ODE's for the curves (the characteristics)  forming the solution, in this case and put in form of proportions:
$$\dfrac{\mathbb dt}{1}=\dfrac{\mathbb dk}{k}=-\dfrac{\mathbb dp}{k^2p}$$
You may prefer to solve $\dfrac{\mathbb dt}{1}=\dfrac{\mathbb dk}{k}$ and $\dfrac{\mathbb dk}{k}=-\dfrac{\mathbb dp}{k^2p}$. Giving two equations with an arbitrary constant for each and defining the characteristics (you can get the general solution from here).
A: If you have never used this method before you need some help setting you on the correct path.
The problem is well posed if we are given initial data of the form $p(k,0) = f(k)$.
With the method of characteristics we find a family of curves $k(t;\alpha)$ that intersects the $k$-axis in the $kt$-plane at a unique point $(\alpha,0)$ and such that the PDE reduces to an ODE along the curves.
The characteristics here solve the IVP,
$$\displaystyle \frac{dk}{dt} = k, \,\, k(0) = \alpha,$$
with solutions $k(t) = \alpha e^t$.
Taking $v(t) = u(\alpha e^t,t)$ you must determine the ODE satisfied by $v$ and solve by applying the intial condition $v(0) = u(\alpha,0) = f(\alpha)$.  Finally transform back to $(k,t)$ coordinates using $\alpha = ke^{-t}$.
This leaves you with a fair amount of work to do.
The ultimate solution will be 
$$ u(k,t) = f(ke^{-t})\exp\left(- \frac{k^2}{2} \right)$$
A: Correction: I thank RRL for catching my mistake. 
Hint: Consider
\begin{align}
u(t, k) = e^{k^2/2}p(t, k)
\end{align}
then check that
\begin{align}
\partial_t u+k\partial_k u = 0
\end{align}
which is just a transport equation.
