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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

  1. Considering a reservoir, with one inlet and one outlet;
  2. And the concentration of a pollutant in the reservoir is $c$;
  3. The concentration of the pollutant in the inlet is a constant $c_i$;
  4. The rate of flow of the inlet and outlet are constants $q_i$ and $q_o$;
  5. 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,

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    $\begingroup$ what all have you tried? $\endgroup$ – user210387 Oct 21 '15 at 6:03
  • $\begingroup$ I don't have much math background, I only know how to solve if the constant 1 is ZERO. $\endgroup$ – Ming Su Oct 21 '15 at 6:12
  • $\begingroup$ simple, variable separable method... $\endgroup$ – Jefree Sujit Oct 21 '15 at 7:02
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    $\begingroup$ In what sense is this a differential equation? What does your equation mean? Manipulating gives $x + y(dx/dy) = dx$, which is not meaningful. Did you make this "equation" up? $\endgroup$ – Michael Oct 21 '15 at 9:28
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    $\begingroup$ Thanks for providing your derivation of the model equation. As others have commented, the equation derived doesn't make sense and cannot be solved. Somewhere you have essentially "divided by zero" without properly compensating for this (e.g. by taking limits), so the real problem here is sorting out your derivation to come up with a meaningful model equation. I'll give it a try. $\endgroup$ – hardmath Oct 21 '15 at 13:00
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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.

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