$u_{xx} + u_x = 0$ ; parabolic PDE

Question

Find the general solution to the equation $u_{xx} + u_x = 0$, assuming that $u$ is a function of two variables, $x$ and $y$

I know this seems like a simple question and I know it has something to do with $e^x$ but I can't figure out a clear way to obtain the general solution

(PS. $u_x$ is the partial derivative of u(x,t) with respect to $x$ and $u_{xx}$ is the second partial derivative)

Thank you in advance!

Answer 1

If $u = u(x)$ you are solving $u'' + u' = 0$ with characteristic equation $0 = a^2+a = a(a+1)$, with roots $0,-1$ and hence a general solution of $$ u(x) = Ae^{0x} + Be^{-1x} = A + Be^{-x}. $$ Since you want $u = u(x,y)$, the form must stay the same, but $A$ and $B$, while constant with respect to $x$, can depend on $y$, so you get $$ u(x,y) = A(y) + B(y)e^{-x}. $$

Answer 2

First, since both derivatives are with respect to x, the fact that u is also a function of y is irrelevant- y just "goes along for the ride". Second, since u itself does not appear, just $u_x$ and $u_{xx}$, I would let $v= u_x$ so that I have the first order equation $v_x+ v= 0$. Then $\frac{dv}{dx}= -v$ so $\frac{dv}{v}= -dx$. Integrating both sides $ln(v)= -x+ C$ so $\frac{du}{dx}= v= C'e^{-x}$ where $C'= e^C$. Integrating again, $u= -C'e^{-x}+ D$.

Since u is a function of y also, but the equation gives no information about y, all we can say is that the "constants", C' and D, may be any functions of y: $$u(x,y)= f(y)e^{-x}+ g(y)$$ where f and g are arbitrary functions.