I want to solve the following partial differential equation:
$$\frac{\partial h}{\partial t} + (y+t)\frac{\partial h}{\partial y} + h = 0$$
So far, the only methods I am familiar with solving PDEs are separation of variables, Laplace transforms, Fourier transforms, change of variables, method of characteristics, and a little bit of Fourier series, though I would like a closed form for the solution. I tried using the method of change of variables to solve the PDE and got the following general answer:
$$h(t,y)=ce^{-t}g(t-\ln\left | y+t \right |)$$
However, I don't think this is correct since when I plugged this in, it turned out to not be the solution. Wolfram Alpha gives the answer as:
$$h(t,y) = ce^{-2t}(t+y+1)$$
From the PDEs I've solved before, it seems obvious that this isn't the general solution. I am pretty sure that Wolfram Alpha chose a specific function $g(c)$, where $c$ is a variable of $t$ and $y$, in order to get a certain solution. Either that or Wolfram chose an initial value to apply. How would I get the general solution to this PDE? Any method would be acceptable. From there, how do you get the specific solution that Wolfram Alpha got? Preferably, I would like to avoid solving the following ODE:
$$\frac{\mathrm{d} y}{\mathrm{d} t}=y+t$$
in the process, since the PDE comes from solving $\frac{\mathrm{d} y}{\mathrm{d} t}=y+t$ as an inexact differential equation. If more work is needed to be shown, I will do so without hesitation.
Best Answer
Suppose $h$ is a solution your original PDE, then consider \begin{align} u(t, y) = e^{t}h(t, y). \end{align} Observe \begin{align} u_t = e^{t}h+e^{t}h_t \end{align} and \begin{align} u_y = e^{t}h_y \end{align} which means \begin{align} u_t+(t+y)u_y = e^t(h_t+(t+y)h_y+h) = 0. \end{align} Hence it suffices to solve \begin{align} u_t+(t+y)u_y = 0. \end{align} which is a good old transport equation.
Using the method of characteristics, observe \begin{align} \frac{d}{dt}u(t, y(t)) = \partial_t u + y'(t)\partial u_y = u_t+(t+y)u_y =0 \end{align} which means \begin{align} y' = t+y \ \ \implies \ \ y(t) = (y_0+1)e^{t}-t-1. \end{align} Hence \begin{align} u(t, (y_0+1)e^{t}-t-1) = \text{ const} = u(0, y_0) = g(y_0) \end{align} which means \begin{align} u(t, y) = g(e^{-t}(y+t+1)-1) \ \ \implies \ \ h(t, y) = e^{-t}g(e^{-t}(y+t+1)-1) \end{align}