# Finding non-homogenous equation of first-order ODE

My professor gave a method for finding the non-homogeneous equation:

$$\frac{dC}{dt}+\frac{F}{V}C=\frac{F}{V}C_0$$

C = concentration in tank
F = flow rate into and out of the tank
V = volume in tank
$$C_0$$ = Concentration of liquid going into the tank

His method is as follows:

Try solution $$C=A$$, constant

-> $$0+\frac{F}{V}A=\frac{F}{V}C_0$$

therefore, $$A=C_0$$

Can somebody please explain to changing 'C' to the constant 'A' leads to the solution for the non-homogeneous part of the equation? Why does this work?

• I'm not quite sure what you're asking. It works in this case due to the particular form of the equation, but in most other cases setting one of the functions to a constant won't work. – John Omielan Aug 2 '19 at 6:44
• I'm asking for a reason why it works in the context of a tank-concentration context.Why would I think "oh yes, lets change the independent variable C to a constant"? – ndyson0 Aug 2 '19 at 6:48
• I gave an explanation in terms of the tank concentration changing but eventually reaching a constant in my answer. – John Omielan Aug 2 '19 at 6:54

If $$a,b$$ are constants and if $$a \ne 0$$, then consider the differential equation
$$(*) \quad y'(t)+ay(t)=b.$$
If $$y_0$$ is the constant function $$y_0(t)=b/a$$, then it is easy to see that $$y_0$$ is a solution of the equation $$(*)$$.
Conclusion: if a differential equation is of the type $$(*)$$, the there exists a solution which is constant.
You can determine this constant $$A$$ by plugging in $$C = A$$ into the differential equation, as you stated. Also, with this current concentration in the tank being a constant, i.e., not changing, then this means that with a flow into and out of the tank occurring, the concentration of the liquid coming into the tank must be the same as this, i.e., equal to that in the tank. Thus, this is an alternate way to see that $$A = C_0$$.