Units of Differential Equations -- MIT Calculus Question

Question

PROBLEM: MIT on this edX course (Unit 1: The integral Differential equations, Section 3. Introducing differential equations ) says units are different for 2 similar kid of differential equations. I can't understand why.

Consider the differential equation $\frac{dx}{dt} = f(t)$ , where $x$ has units of meters, $t$ and has units of seconds

REASONING #1: The functions $x = \int f(t)dt$ are the solution curves to the differential equation $\frac{dx}{dt} = f(t)$. Thus $\int f(t)dt$ must have the same units as $x$, which are meters.

The symbol $dt$ stands for the infinitesimal version of $\Delta t$, which is a tiny bit of $t$, thus it has the same units as $t$, which is seconds.

The differential quantity $f(t)dt = dx$, so $f(t)dt$ has the same units as $dx$, which is a little bit of $x$, and must have units of meters.

Here is the 2ns question:

REASONING #2: The units of $y$ are the units of $f(x)$ times the units of $dx$ , which are the same as the units of $x$ . This gives units of meters squared, which is a measure of area.

It contradicts with what they said earlier. I can not understand this, for $\int f(t)dt$ units are meters but for $\int f(x)dx$ units are meters squared. They both looks same to me, why then same kind of symbols have different kinds of units. Even if I take $\int f(u)du$ or $\int f(h)dh$, it should not change the meaning of Mathematical Symbols.

Answer 1

You know $f(x)$ and $x$ has units of meter. So $\frac{dy}{dx}$ is $\frac{dy}{meter}=f(x)=meter$

Thus solving for $dy$ you get ${meter}^2$.

Answer 2

The difference is that in the first case $f(x)$ has units meters per second (since it equals the velocity $dx/dt$).

Answer 3

You may just setup an equation for the units using the common symbol $[Q]$ as the unit of a quantity $Q$.

• $\frac{dx}{dt} = f(t)$: $[f(t)] = [\frac{dx}{dt}] = \frac{[dx]}{[dt]}=\frac{m}{s} \Rightarrow [x]=\left[\int f(t)\,dt \right] = \frac{m}{s} \cdot s = m$

• $\frac{dy}{dx} = f(x)$: $[f(x)] = [\frac{dy}{dx}] = \frac{[dy]}{[dx]}=\frac{[dy]}{m}=m \Rightarrow [y]= \left[\int f(x)\,dx \right] = m \cdot m = m^2$