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I am studying differential equations right now and I am confused the way $dx$ is being used.

When I learnt calculus I thought that $\frac{df}{dx}$ is just symbolic representation of derivative and we can't use it as a fraction. But when it comes to differential equation like $y = \frac{df}{dx}$ they do the following:

$$ydx = df$$

$$\int ydx = f$$

And then again I was taught that $dx$ in integral has no meaning but only convenient way to represent integral.

I am totally fine with the intuition why they do that, I can imagine $\frac{df}{dx}$ meaning small change in $f$ divided by small change in $x$ and integral meaning sum of recatngles of width $dx$ and height $f$, but what I lack is rigorous transition from $\frac{df}{dx}$ meaning just a symbolic representation of derivative to the state where we can algebraically manipulate it as a fraction. Could you please help me with that or suggest some reading?

Also, could you please recommend me some books on differential equations that teach intuition behind the equations?

Rodrigo de Azevedo
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Studying Optimization
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2 Answers2

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Yes, you may get into trouble if you see $\frac{\mathrm dy}{\mathrm dx}$ as a fraction. But the reason why we do what in described in the context of differential equations is that it works. If we have the differential equation $\frac{\mathrm dy}{\mathrm dx}=f(y)$, what we do is$$\frac{\mathrm dy}{\mathrm dx}=f(y)\iff\frac{\mathrm dy}{f(y)}=\mathrm dx\implies\int\frac{\mathrm dy}{f(y)}=\int\mathrm dx.$$So, what we do is:

  • take a primitive $\varphi$ of $\frac1f$
  • a solution of your differential equation will be $\varphi^{-1}$.

And this works because$$(\varphi^{-1})'(x)=\frac1{\varphi'\bigl(\varphi^{-1}(x)\bigr)}=f\bigl(\varphi^{-1}(x)\bigr),$$So, yes, $\varphi^{-1}$ is indeed a solution.

So, the method works and, if we see $\frac{\mathrm dy}{\mathrm dx}$ as a fraction, it is easy to memorize.

I suggest that you read Wolfgang Walter's Ordinary Differential Equations.

José Carlos Santos
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You may see $d(...)$ as an operator on $f$ defined by $d(f(x))=f'(x)dx$, where $dx$ is a symbol meaning that we compute the derivative with respect to $x$. Note that this defenition stays consistent with $f(x)=x$ since then $d(x)=1\cdot dx$. In addition, if $y=f(x)$, then $dy=f'(x)dx=y'dx$.

This point of view gives us usefull and natural identities, i.e. $$ d(xy)=(xy)'dx=(xy'+x'y)dx= x(y'dx)+ydx=xdy+ydx $$

boaz
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