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If I want to solve an equation like $x' = -2t(x+1)$, I will write $\dfrac{dx}{dt} = x'\;\;\implies\;\;\dfrac{dx}{dt} = -2t(x+1)$ and then I can solve it for $x$.

But why is it possible to write $\dfrac{dx}{dt}$ instead of $x'$?

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    $f$ixed tag: not (difference-equations)2012-12-29

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...why is it possible to write $\dfrac{dx}{dt}$ instead of $x'$?

$x' = -2t(x+1)\iff \frac{dx}{dt} = -2t(x+1)$

I.e., $\;x'\;$ is "shorthand" for $\;\dfrac{dx}{dt}$.

What's nice about $\;\dfrac{dx}{dt}\;$ is that it specifies the derivative of $\;x\;$ with respect to $\;t$.
But the same is taken as given when you see $\;x' = -2t(x+1),\;$ which is more brief than using $\;\dfrac{dx}{dt} = -2t(x+1)$.

If you also have $y' = at(y + b)$ where $a, b$ are given, you could write $\;\dfrac{dy}{dt} = at(y + b)$. If you know $\;\dfrac{dy}{dt}\;$ and $\;\dfrac{dx}{dt},\;$ you can solve for $dy/dx$:

$ \begin{align*} \dfrac{dy}{dx}\dfrac{dx}{dt} &= \dfrac{dy}{dt} \\ \\ \text{So}\;\;\;\dfrac{dy}{dx} &= \dfrac{dy/dt}{dx/dt}\end{align*}$