differential equations - change of unit

Question

$$ \frac {d y}{d t} = -0.002\, y $$ In the above differential equation y is the amount of salt in kg and t is the time in minuttes. I was wondering when i solve the equation whether the time will change unit or not? I can solve the equation with the method separation of variables. If i integrate will the time then change unit to seconds?

Answer 1

The left hand side of your equation has dimension $kg \over min$. The right hand side must have the same dimension. But $y$ has dimension $kg$ which means that constant 0.002 must have dimension $1 \over min$.

In other words your equation looks like this:

$$\frac {d y}{d t} = -0.002\frac{1}{min}\cdot y$$

Solve equation normally. You get:

$$ \ln{y}=-0.002\frac{1}{min}\cdot t+ln{C}$$

...or:

$$ y = C e^{-0.002\frac{1}{min}\cdot t}$$

For any given $t$ just make sure that you cancel the units in the exponent properly. For example, for $t=30s$ the value in the exponent would be:

$$ -0.002 \frac{1}{60s} \cdot 30s = -0.001$$

Just have in mind that in the physical world most constants in differential equations have hidden dimensions. You have to understand those dimensions properly, otherwise you won't be able to calculate anything properly.

Answer 2

If you change variables $s=60t$ (when $t=1$ min then $s=60$ sec) to get a time in seconds.

You have $\mathop{ds}=60\mathop{dt}$

And the equation becomes $\dfrac{\mathop{dy}}{\mathop{dt}}=\dfrac{\mathop{dy}}{\frac 1{60}\mathop{ds}}-0.002y\implies \dfrac{\mathop{dy}}{\mathop{ds}}=-\dfrac{y}{30000}$

Answer 3

Why would the units change? A unit should be treated as a constant for the purposes of this equation. So:

$$\begin{align} \frac{\mathrm{d}y}{\mathrm{d}{t}}&=-0.002y\cdot\text{min}^{-1}\\ \mathrm{d}y&=-0.002y\cdot\text{min}^{-1}\,\mathrm{d}t\\ \frac{\mathrm{d}y}{y}&=-0.002\cdot\text{min}^{-1}\,\mathrm{d}t\\ \int\frac{\mathrm{d}y}{y}&=\int-0.002\cdot\text{min}^{-1}\,\mathrm{d}t\\ \ln|y|&=-0.002\cdot\text{min}^{-1}\cdot t+C\\ y&=\pm e^{-0.002\cdot\text{min}^{-1}\cdot t+C}\\ y&=C\cdot e^{-0.002\cdot\text{min}^{-1}\cdot t} \end{align} $$ The minutes are still there, and will cancel with the unit of $t$.