How to solve the pde $\frac{\partial^3f}{\partial x\partial y\partial z}=f(x,y,z)$. How to solve the pde $\frac{\partial^3f}{\partial x\partial y\partial z}=f(x,y,z)$.
I know just $f'=f$. What if the problem?
 A: As AHusain mentions, the independence of $x,y,z$ suggests a solution of the form $f(x,y,z)=X(x)Y(y)Z(z)$. Since the PDE is first-order with respect to each variable, try a solution of the form $X(x)=C_1e^{a_1x},Y(y)=C_2e^{a_2y},Z(z)=C_3e^{a_3z}$.
Then: $f(x,y,z)=C_1e^{a_1x}C_2e^{a_2y}C_3e^{a_3z}=Ce^{a_1x+a_2y+a_3z}$.
Taking the partial derivative of this, we get
$\frac{\partial^3}{\partial x\partial y\partial z}[Ce^{a_1x+a_2y+a_3z}]=
\frac{\partial^2}{\partial x\partial y}[a_3Ce^{a_1x+a_2y+a_3z}]=
\frac{\partial}{\partial x}[a_2a_3Ce^{a_1x+a_2y+a_3z}]=
a_1a_2a_3Ce^{a_1x+a_2y+a_3z}$.
Plugging this into our PDE, we get:
$a_1a_2a_3Ce^{a_1x+a_2y+a_3z}=Ce^{a_1x+a_2y+a_3z}$
$\to a_1a_2a_3=1$
Solving for $a_3$, we get $a_3=\frac{1}{a_1a_2}$
Thus any function of the form $f(x,y,z)=Ce^{ax+by+\frac{z}{ab}}$ should solve the PDE.
A: In order to complete the answer from ash4fun :
Any linear combination of particular solutions is solution of the PDE.
$$f(x,y,z)=\sum_{\forall a} \sum_{\forall b} C_{a,b}e^{ax+by+\frac{1}{ab}z}$$
$C_{a,b}$ are arbitrary constants.
Or instead of the above discrete form :
$$\boxed{f(x,y,z)=\int\int \varphi(a,b)\:e^{ax+by+\frac{1}{ab}z}\:da\:db}$$
$\varphi(a,b)$ is any arbitrary function of two variables.
