Number of arbitrary constant in a Partial differential equation Recently, I have been taught that second order Ordinary differential equation must have two arbitrary constants, but is it true that for a Partial differential equation PDE, with two variables x,y they should have 4 arbitrary constants?
How can we predict the number of initial conditions or boundary conditions for a certain PDE partial differential equation problem?
 A: As we move  from ODE to  PDE, the solution space becomes infinite dimensional. Consider the very simple PDE $$\dfrac{\partial u}{\partial x}=0\quad \text{in } \ \mathbb R^2$$ which has solutions $u(x,y)=g(y)$ for $g$ an arbitrary function. Similarly, the equally simple PDE
$$\dfrac{\partial^2 u}{\partial x^2}=0\quad \text{in } \ \mathbb R^2$$ 
has solutions $u(x,y)=g(y)+xh(y)$ for two arbitrary functions $g$ and $h$. You can see a pattern here: the number of initial conditions   is the order of partial derivative that is transverse (i.e., not parallel)  to the line (or curve, or surface) on which we prescribe the initial  condition. 
The above pattern holds in many cases, but it should not be taken as an absolute truth. Basically, it holds for evolution-type equations (wave or diffusion) which can be recast as a ODE in some function space. For example, the heat equation $u_t=k\Delta u$ is a first-order ODE in this sense, if we think of $u(x,t)$ as a function $t\mapsto u$ where $u$ is an element of some function space defined on a domain in $x$-space. 
Here is a tricky one: how many conditions on the axis $y=0$ can we impose on solutions of the Laplace equation $u_{xx}=u_{yy}$? On one hand, we can fulfill two conditions $u(x,0)=g(x)$ and $u_y(x,0)=h(y)$ if $g$ and $h$ are very nice (real analytic). On the other hand, the single condition $u(x,0)=g(x)$  determines a unique solution with desirable properties in upper halfplane, where $g$ no longer needs to be nice.
A: The number of constants of integration should equal the number of derivatives. For instance
$$
\frac{\partial^2 f(x)}{\partial x^2}=0
$$
Would need two boundary conditions (or equivalently would have two constants of integration), likewise
$$
\frac{\partial^2 f(x, t)}{\partial x^2}=\frac{\partial f(x, t)}{\partial t}
$$
Would have need three boundary conditions to get an explicit solution. As you mentioned, equations like
$$
\frac{\partial ^2 f(x, y)}{\partial x^2}+\frac{\partial^2 f(x, y)}{\partial y^2}=0
$$
Would have 4 constants of integration. Or if solving the equation in real life, you would need 4 defined boundary conditions, (the upper/lower bound of $x$ and $y$),  to solve for the constants of integration.
