Does solve PDE by combination of variables always cannot find the general solutions? Combination of variables is the technique that reducing the PDE to one independent variables (i.e. become ODE) by introducing a suitable change of variables.
But the general solution of a PDE should contain arbitrary functions as the arbitrary parts, while the general solution of an ODE only have arbitrary constants as the arbitrary parts, does it mean that solving PDE by combination of variables always cannot find the general solutions?
Saying typically, combination of variables is the technique described in the following references:


*

*Combination of Variable Solutions

*Solution of Partial Differential Equations - Combination of Variables

*Second-order PDE
For the example in Combination of Variable Solutions, it claims that solving $\dfrac{\partial\theta}{\partial t}=\gamma\dfrac{\partial^2\theta}{\partial x^2}$ with $t>0$ , $0<x<\infty$ , $\theta(x,0)=\theta_0$ , $\theta(0,t)=\theta_1$ , $\theta(\infty,t)=\theta_0$ by combination of variables is $\theta(x,t)=\theta_1+(\theta_0-\theta_1)\text{erf}\left(\dfrac{x}{\sqrt{4\gamma t}}\right)$ .
I discover that when solving the same problem by separation of variables, i.e. Random diffusion coefficient in the Fourier equation with replacing $u$ by $\theta$ , $u_0$ by $\theta_1$ , $u_L$ by $\theta_0$ , $k(t)$ by $\gamma$ , the solution is $\theta(x,t)=\theta_1+\int_0^\infty\dfrac{2(\theta_0-\theta_1)e^{-\gamma ts^2}\sin xs}{\pi s}ds=\theta_1+(\theta_0-\theta_1)\text{erf}\left(\dfrac{x}{\sqrt{4\gamma t}}\right)$ . This seems show that solving PDE by combination of variables takes the same effect to solving PDE by separation of variables.
But when the problem is replacing to $\dfrac{\partial\theta}{\partial t}=\gamma\dfrac{\partial^2\theta}{\partial x^2}$ with $t>0$ , $0<x<\infty$ , solve it by combination of variables Combination of Variable Solutions claims that $\theta(x,t)=I_1\text{erf}\left(\dfrac{x}{\sqrt{4\gamma t}}\right)+I_2$ , but solve it by separation of variables Random diffusion coefficient in the Fourier equation claims that $\theta(x,t)=C_1x+C_2+\int_0^\infty C_3(s)e^{-\gamma ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-\gamma ts^2}\cos xs~ds$ . This seems that knowing whether solving PDE by combination of variables and solving PDE by separation of variables take the same effect or not are difficult.
So how should I recognize those issues? In fact $I_1$ and $I_2$ are arbitrary constants or arbitrary functions? If they are arbitrary functions, i.e. $I_1=I_1(f(x,t))$ and $I_2=I_2(g(x,t))$ , $f(x,t)$ and $g(x,t)$ are equal to what?
 A: What you call "combination of variables" I think is often called "separation of variables" in my humble sphere of math. Let's take an example:
$$u_{xx}+u=u_t$$
Here, $u$ is the dependent variable and $x,t$ are independent. Usually, we suppose $u(x,t) = X(x)T(t)$ hence $u_{xx} = X''T$ and $u_t = XT'$. Substituting into the PDE gives,
$$X''T+XT=XT' $$
If the solution we seek is nontrivial then division by $XT$ is allowed,
$$ \frac{X''}{X}+1 = \frac{T'}{T} $$
At this point we note that the left and right sides of this equation are functions
of variables which we took as independent. It follows there is some constant $K$ such that:
$$ \frac{X''}{X}+1 = \frac{T'}{T} =K$$
Hence, we find solving the PDE reduces to solving two ODEs:
$$ T' = KT \qquad X''+(1-K)X=0. $$
Notice that $K$ is arbitrary with the data given. This means that we don't
really have just two ODES, we have infinitely many. What typically happens is that
in a well-posed problem you are also given Boundary Values (BV) and an initial condition.
In nice problems the BVs will force $K$ to be a certain discrete set of values, countably many in fact. Then for each choice of $K$ you get two ODEs and a couple of solutions to the common BV. After all of this you sum these together to form the formal solution and match it to the initial condition(s) by the Fourier technique. So, in that sense there are not just a couple constants as in the solution to a single ODE. There is infinite flexibility in the solution because you have a family of ODEs forming the solution. 
