Difficulty solving equation with partial derivatives So when I was in school I never went past college algebra.  But I have encountered a specific equation which I want to understand.  
At first I thought that an afternoon's focus would be enough to crack it, but no, it was too hard.  I have stuck at it for four days now.  This is the equation:
$\lambda^2\dfrac{\partial^2 V'}{\partial x^2}+t\dfrac{\partial V'}{\partial t}+V' = 0$
The partial derivatives throw me.  Am I understanding correctly that the partial derivative is basically a function that evaluates to a variable value?  
Following that presumption, I searched the book I'm reading and found the following:
$\dfrac{\partial^2 V'}{\partial x^2} = V'\dfrac{r_i + r_o}{r_m}$
Thinking I know how to solve that, I did.  I'll call the result $x$
Then I find the following:
$\dfrac{1}{r_i + r_o}\dfrac{\partial^2 V'}{\partial x^2} = \dfrac{V'}{r_m} + c_m\dfrac{\partial V'}{\partial t}$
I think, that looks like what I need, if I isolate it ... So I did this:
$((\dfrac{1}{r_i + r_o}\dfrac{\partial^2 V'}{\partial x^2})/c_m) - \dfrac{V'}{r_m} = \dfrac{\partial V'}{\partial t}$
Then I solved that.  I'll call the result $y$
Finally, I returned to the main problem, which I solved like this:
$\lambda^2x+ty+V' = 0$
Except, it didn't equal zero.  Did I get the concepts correct?
 A: This is about an unknown function $(x,t)\mapsto U(x,t)$ (called $V'$ in your question) whose domain $\Omega$  is (part of) the $(x,t)$-plane. Consider a point $(x_0,t_0)\in\Omega$. If you look at this function along the horizontal line $t=t_0$  then you obtain a function
$$\phi(x):=U(x,t_0)$$
of the single variable $x$, which presumably has a derivative at $x_0$. This derivative is called the partial derivative of $U$ with respect to $x$ at $(x_0,y_0)$, and is denoted by $${\partial U\over\partial x}(x_0,t_0)=U_x(x_0,t_0)\ .$$ 
The partial derivative
$${\partial U\over\partial t}(x_0,t_0)=U_t(x_0,t_0)$$ 
is defined similarly. 
The PDE you are given seems to describe some "law of nature" about the unknown function $U$, insofar as the partial derivatives $U_{xx}$ (differentiate twice with respect to $x$) and $U_t$ are related in a particular way:
$$\lambda^2U_{xx}+tU_t+U=0\qquad\forall (x,t)\in\Omega\ .$$
In order to simplify matters I'm going to replace the variable $t$ by a new variable $\tau$ connected to $t$ via $t:=e^\tau$. Since by the chain rule $$U_\tau= U_t\cdot{dt\over d\tau}= U_t e^\tau=t\>U_t\ ,$$ we therewith obtain a constant coefficient equation in the new variables $(x,\tau)$:
$$\lambda^2U_{xx}+U_\tau+U=0\qquad\forall (x,\tau)\in\Omega'\ .\tag{1}$$
The standard recipe for $(1)$ is separation of variables: One is looking for a bag full of  solutions of $(1)$ of the special form
$$U(x,\tau)=X(x)\cdot T(\tau)\ .$$
But this is another matter.
