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Let's suppose we have an ODE of the form $$x^\prime=f(x) $$ for some convenient function $f:\mathbb{R}\to\mathbb{R}$.

Suppose we don't know the Leibniz notatin $\dfrac{dx}{dt}=x^\prime$. Ok, now how do we solve the above ODE without using the suggestive Leibniz notation? Because using his notation we have $\dfrac{dx}{dt}=f(x)$ and magically we can deduce that $dx/f(x)=dt$ ($f\neq 0$) and intagrate in both sides and we will have a solution. But why this works? I'm trying to find out why this really works. Let's do some other example. Consider the ODE $$x^\prime=\dfrac{f(x)}{g(t)}$$ for some convenient functions $f$ and $g$ and suppose we can solve analytically this ODE. So, how to do that? I cannot do this task whithout the Leibniz notation ($\dfrac{dx}{dt}=f(x)/g(t)$ implies $dx/f(x)=dt/g(t)$ so $\int 1/f(x)dx=\int(1/g(t)dt)$). How can I find out the same result whithout using the arguments I used in the parentesis? I do not know if I'm being clear, but what I want is to really understand what I'm doing on solving an ODE. Any help will be appreciated, thanks.

Jacaré
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1 Answers1

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The method of separation of variables is legitimate, regardless of the notation you employ. The idea is that the implicitly defined equation $F(x) = G(t) + C$ where $$ F'(x) = \frac{1}{f(x)} \qquad\text{and}\qquad G'(t) = \frac{1}{g(t)} $$ describes a curve that satisfies the differential equation $$ x'(t) = \frac{f(x)}{g(t)}. $$

Why? Differentiate both sides of the solution equation with respect to $t$ and remember that $x$ is implicitly a function of $t$, so we need the Chain Rule! \begin{align} &&\frac{d}{dt} \bigl( F(x) \bigr) &= \frac{d}{dt} \bigl( G(t) + C \bigr) \\ &\implies & F'(x) \, x'(t) &= G'(t) \\ &\implies & \frac{x'(t)}{f(x)} &= \frac{1}{g(t)} \\ &\implies & x'(t) &= \frac{f(x)}{g(t)} \end{align}

Sammy Black
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