How many "constants of integration" will a PDE have?

To address your first question, it's important to realise that you can write any $n$-th order ODE as a first order ODE in $n$ variables, introducing $$ y_0 = x,\quad y_1 = \frac{\text{d} x}{\text{d} t},\quad y_2 = \frac{\text{d}^2 x}{\text{d} t^2},\quad (\ldots) ,\quad y_{n-1} = \frac{\text{d}^{n-1} x}{\text{d} t^{n-1}}. $$ Using these definitions, you can write a general $n$-th order ODE $$ \sum_{i=0}^n a_i(t) \frac{\text{d}^i x}{\text{d} t^i} = b(t) $$ as $\frac{\text{d}\mathbf{y}}{\text{d} t} = F(\mathbf{y},t)$, with $\mathbf{y} = (y_0,\ldots,y_{n-1})$ and $$ F(\mathbf{y},t) = \left(y_1,y_2,\ldots,y_{n-1},-b(t)-\sum_{i=0}^{n-1} a_i(t) y_i \right), $$ where I assumed that $a_n(t) = 1$ without too much loss of generality. Now, by the local existence and uniqueness theorem for ($n$-dimensional) ODEs, we know that for a local time interval $t \in (a,b)$, the solution to the ODE $\frac{\text{d}\mathbf{y}}{\text{d} t} = F(\mathbf{y},t)$ is uniquely determined by its initial condition $\mathbf{y}(0) = \mathbf{y_0}$ (see e.g. the Picard-Lindelöf theorem). So, to uniquely determine a solution to an $n$-dimensional ODE, you can pick an $n$-dimensional vector $\mathbf{y}_0$ of your choice. The $n$ components of the vector $\mathbf{y}_0$ play the role of the $n$ 'constants of integration'.

For PDEs, the situation is much more complex. This is because, whereas an ODE defines a flow in an $n$-dimensional phase space, a PDE defines a flow in a function space. I'm skipping a lot of very important and interesting details here, but the important thing is that function spaces are infinite dimensional. To make the connection with the ODE case, take the example of the advection equation, given by $$ c \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} = 0. $$ You can use the method of characteristics to write down the general solution to this PDE. The solution is given by $$ u(x,t) = f(x - c t), $$ where $f(x)$ is a function that plays the role of the initial condition, since $u(x,0) = f(x)$. So, you can pick any function $f(x)$ you like (again, skipping a couple of important and interesting details here), and you get a solution to the PDE. As there are, roughly speaking, infinitely many functions to choose from, you have infinitely many 'degrees of freedom' in picking an initial condition (initial function) for your PDE.

The order of the PDE mainly influences the particular function space it acts on, so the function space in which you can pick a function as initial condition. However, typically, all these function spaces are infinite dimensional -- in the sense that they are not finite-dimensional. The field of functional analysis addresses questions that arise in this context.

The pde $-\Delta u = f$ has many 'constants of integration': you can add any harmonic function $v$, i.e., $-\Delta v=0$, to a solution $u$ of the pde above. There are infinitely many such functions