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I'd like to solve the following differential equation: $$\frac{dQ}{dt}=\frac{k_BT}{m}-\frac{\alpha Q}{m}$$ where $Q$ has units of $\text{m}^2\text{s}^{-1}$, $k_B$ is Boltzmann's constant, $T$ is temperature, $m$ is a mass and $\alpha$ has units $\text{kg s}^{-1}$.

Separating variables gives me $$\int\frac{dQ}{\frac{k_BT}{m}-\frac{\alpha Q}{m}}=\int dt$$ Hence $$-\frac{m}{\alpha}\ln\left(\frac{k_BT}{m}-\frac{\alpha Q}{m}\right)=t+C$$

Now, all of a sudden, I have something which does not hold up when I look at the dimensions: taking the logarithm of a dimensionfull quantity does not make any sense. Multiplying by $-\frac{\alpha}{m}$ and exponentiating yields $$\frac{k_BT}{m}-\frac{\alpha Q}{m}=e^{-\frac{\alpha t}{m}-C'}=C''e^{-\frac{\alpha t}{m}}$$ which is again problematic (left hand side has units $\text{m}^2\text{s}^{-2}$ while right hand side is unitless).

My question is: what am I missing here? This technique of solving ODE's is widely used, so I must be overlooking something simple. However, my professor was equally stumped as I am.

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  • $\begingroup$ The thing that saves you is the integration constant. You can write it as $C = m/\alpha \ln X$ for some unknown $X$. Then you can bring the logs together and if you choose $X$ to have the correct dimensions, your log becomes dimensionless. But in general it is better to make everything dimensionless before solving the ODE. $\endgroup$ – Vibert Oct 3 '13 at 12:21
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When using this method of solving ODEs, we have to be careful not to use the indefinite integral.

$\int\frac{dQ}{\frac{k_BT}{m}-\frac{\alpha Q}{m}}=\int dt$ is actually $\int\limits_{Q_0}^Q\frac{dQ}{\frac{k_BT}{m}-\frac{\alpha Q}{m}}=\int\limits_{t_0}^t dt$

This gives us

$$-\frac{m}{\alpha}\ln\left(\frac{k_BT}{m}-\frac{\alpha Q}{m}\right)+ \frac{m}{\alpha}\ln\left(\frac{k_BT}{m}-\frac{\alpha Q_0}{m}\right)=t-t_0$$

As $\ln a - \ln b = \ln \frac{a}{b}$, this can be rewritten as

$$-\frac{m}{\alpha}\ln\left(\frac{k_BT-\alpha Q}{k_BT-\alpha Q_0}\right)=t-t_0$$

Which has a dimensionless argument of the logarithm.

Another way of looking at it is that the constant $c$ you got after taking the indefinite integral was of the form $c_0-\ln(c_1)$, where $c_1$ has the same units as that of $\frac{\alpha Q}{m}$.

Usually, in mechanics (and most of physics), integrals are definite, even if we don't write them that way.

Also,while solving ODEs, it's a good idea to normalize everything to a dimensionless quantity before proceeding to solve. This isn't such a big deal for simple ODEs like the one above, but when dealing with complex coupled systems losing the units leads to fewer things to keep track of.

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