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For years I've heard that differentials aren't fractions and multiplying $\frac {d y}{d x} $ with $d x$ to "cancel out the denominator" isn't the right view, yet I've never been shown this "right view" and in my ODE class we always do exactly what we were told was wrong.

What exactly are the properties and proper usages of differentials? How come we're told they don't technically "cancel out" in fractions, yet nearly everything we've been exposed to counters that?

Andrew Tawfeek
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  • Related, possibly duplicates: http://math.stackexchange.com/questions/1252405/is-it-mathematically-valid-to-separate-variables-in-a-differential-equation and http://math.stackexchange.com/questions/27425/what-am-i-doing-when-i-separate-the-variables-of-a-differential-equation – symplectomorphic Mar 11 '17 at 07:51
  • Also: http://math.stackexchange.com/q/1356981/34930 – celtschk Mar 11 '17 at 10:02

2 Answers2

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Consider the following example \begin{align} \frac{dy}{dx} = xy \end{align} which is a separable differential equation. In particular, it is taught in most ODE courses that you could rewrite the equation as \begin{align} \frac{1}{y}dy = xdx \end{align} then integrate both sides accordingly and everything will workout. However, a more rigorous way is to view as follow \begin{align} \int \frac{1}{y}\frac{dy}{dx}\ dx = \int x\ dx \end{align} then by the change of variable formula we get that \begin{align} \int\frac{1}{y} \frac{dy}{dx}\ dx = \int \frac{1}{y}\ dy. \end{align} So by treating the ordinary derivative as a fraction, you simply save the step for justifying the change of variable.

Jacky Chong
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  • Is this always the case though? Is there a case where treating differentials as fractions could bring about errors in calculation? – Andrew Tawfeek Mar 11 '17 at 07:15
  • Remember, $\frac{dy}{dx} \approx \frac{\Delta y}{\Delta x}$ which is a fraction. However the only difference is that there is also a limit involved in the definition of derivative which means that you might have infinity over infinity as a fraction. – Jacky Chong Mar 11 '17 at 07:19
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    Your use of the word "then" in "then by the change of variables formula" is confusing: it suggests, or at least could suggest, the equation that follows is deduced from the previous equation, but it's not. (I would have said "but the change of variables formula says the lefthand side is $\int\frac{1}{y},dy$.") – symplectomorphic Mar 11 '17 at 07:48
  • @AndrewTawfeek : Yes, there are restrictions on the substitution that the "short way" may occlude: The substitution is only valid on monotone branches of the substitution function, i.e., where it is bijective. Any extension beyond must be justified and may require the adjustment of integration constants – Lutz Lehmann Mar 11 '17 at 09:37
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Given Co-Ordinate projections in ${\mathbb{R}^2}$

\begin{gathered} x:{\mathbb{R}^2} \to \mathbb{R},x\left( p \right) = u \hfill \\ y:{\mathbb{R}^2} \to \mathbb{R},y\left( p \right) = v \hfill \\ \end{gathered}

point $p \in {\mathbb{R}^2}$ can now be written like

$$p = \left( {u,v} \right)$$.

Differentials from these Co-Ordinate functions at point $p$

\begin{gathered} d{x_p} = {T_p}{\mathbb{R}^2} \to \mathbb{R} \hfill \\ d{y_p} = {T_p}{\mathbb{R}^2} \to \mathbb{R} \hfill \\ \end{gathered}

are Co-Vectors, that means $d{x_p},d{y_p} \in T_p^*{\mathbb{R}^2}$.

For every Tangent-Vector like $\xi \in {T_p}{\mathbb{R}^2}$, we have

$$\xi = a\frac{\partial }{{\partial x}} + b\frac{\partial }{{\partial y}}$$

$a,b \in \mathbb{R}$. Now

$$d{x_p}\left( \xi \right) = d{x_p}\left( {a\frac{\partial }{{\partial x}} + b\frac{\partial }{{\partial y}}} \right) = d{x_p}\left( {a\frac{\partial }{{\partial x}}} \right) + d{x_p}\left( {b\frac{\partial }{{\partial y}}} \right) = a\frac{{\partial x}}{{\partial x}} + b\frac{{\partial x}}{{\partial y}} = a$$

$$d{y_p}\left( \xi \right) = d{y_p}\left( {a\frac{\partial }{{\partial x}} + b\frac{\partial }{{\partial y}}} \right) = d{y_p}\left( {a\frac{\partial }{{\partial x}}} \right) + d{y_p}\left( {b\frac{\partial }{{\partial y}}} \right) = a\frac{{\partial y}}{{\partial x}} + b\frac{{\partial y}}{{\partial y}} = b$$.

That is: $d{x_p},d{y_p} \in T_p^*{\mathbb{R}^2}$ are linear real-valued functions. And from this point of view we are calculating with real numbers.

So we can divide them like $\frac{{d{y_p}}}{{d{x_p}}}$ and there is nothing to worry about. If $f:{\mathbb{R}^2} \to \mathbb{R},y = f\left( x \right)$ for a differentiable function, we have

$$\frac{{d{y_p}}}{{d{x_p}}} = f'\left( p \right)$$

Nothing, but a real number.

Frieder
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