How to solve this differential equation？ x/dx+y/dy=1 How to solve this differential equation?
$\frac{x}{dx}+\frac{y}{dy}=1$
Thank you very much.
Ok, let me say little more about the meaning of this equation.
water quality dynamic model


*

*Considering a reservoir, with one inlet and one outlet;

*And the concentration of a pollutant in the reservoir is $c$;

*The concentration of the pollutant in the inlet is a constant $c_i$;

*The rate of flow of the inlet and outlet are constants $q_i$ and $q_o$; 

*The water volume of reservoir is $V$, with the initial volume of $V_0$.


Then we want to resolve the concentration $c$ in the reservoir with time $t$.
$c+\Delta c=\frac{V\times c+q_i\times \Delta t\times c_i-q_o\times \Delta t\times c}{V+\Delta t\times(q_i-q_o)}$
$V=V_0+(q_i-q_o)\times t$
==>
$\frac{V_0+(q_i-q_o)\times t}{\Delta t}+\frac{q_i(c-c_i)}{\Delta c}=q_i-q_o$
it just like:
$\frac{f(t)}{dt}+\frac{g(c)}{dc}=C$
Any where wrong?
Thanks,
 A: If you let $P(t) = V(t)c(t)$ be the pollutant size at time $t$, and if you assume $V(t) > 0$ for all $t \geq 0$, then you can write: 
$$ q_ic_i - q_oc(t) = P’(t) = V’(t)c(t) + V(t)c’(t) $$
Since $V(t)$ is a known function, this simplifies to a standard linear first-order ODE that can be solved. 

If you want to derive the above equation from "small changes" you would write: 
$$ (q_ic_i - q_o c(t))(\Delta t) = \Delta P $$ 
then divide by $\Delta t$ and let $\Delta t \rightarrow 0$, noting that $\lim_{\Delta t\rightarrow 0} \frac{\Delta P}{\Delta t} = P'(t)$.  
With alternative notation, you let $h>0$ be a small value and write:
$$  (q_ic_i-q_0c(t))h \approx P(t+h)-P(t)  $$
Then divide by $h$ and use $\lim_{h\rightarrow 0} \frac{P(t+h)-P(t)}{h}=P'(t)$. 

For your equation, I cannot follow the arithmetic since the notation is awkward and my $c_i$ and $c$ and $q_i$ and $q_o$ keep getting mixed up.  Anyway, your original equation is something like: 
$$ c + \Delta c = \frac{Vc + q_i (\Delta t)c_i -q_o (\Delta t)c}{V + (\Delta t)(q_i-q_o)}$$ 
I understand this equation, but I do not know why you would want to write such an equation. Anyway, as @hardmath suggests, when taking a limit it does not give anything meaningful (we want a derivative to pop out in the limit).  When taking $\Delta t \rightarrow 0$ we just get $\Delta c \rightarrow 0$ and the equation reduces to the (unhelpful) equation $c=c$.
With alternative notation, you equation becomes (with $h>0$ a small value): 
$$ c(t+h) \approx \frac{Vc(t) + q_ic_i h - q_0c(t)h}{V(t+h)}$$
Taking a limit now as $h\rightarrow 0$ does not help (it gives $c(t)=c(t)$). You could subtract $c(t)$ from both sides, divide both sides by $h$, and then take a limit as $h\rightarrow 0$. Then the left-hand-side legitimately converges to $c'(t)$ and you end up with the same ODE as I give above. 
