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I'm reading about ODE's in a calculus text. The goal is to solve the differential equation: $\frac{dx}{dt}=ax$ for $x(t)$. The text says using the chain rule and substitution rewrite the equation as $\frac{dx}{x}=a\:dt$.

I'm not sure how to use the chain rule to do the rewriting. However after this step I was able to follow to the solution.

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  • $\begingroup$ Perhaps they are thinking of $$dt = \frac{dt}{dx} dx$$ as an expression of the chain rule. $\endgroup$ – WW1 Dec 19 '20 at 5:09
  • $\begingroup$ @WW1 yes I think that's what they meant if: given $\frac{dx}{dt}$ then $dx=\frac{dx}{dt}dt$. $\endgroup$ – Alexander Orman Dec 19 '20 at 5:36
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To see how this works in a general context, observe that the first order differential equation is separable and is of the form

$$ \frac{dx}{dt}=F(t)G(x). $$ To solve this equation we can divide by $G(x(t))$ to get $$\frac{1}{G(x(t))}\frac{dx}{dt}=F(t).\tag{1}$$ Then we can find a function $H(x)$ whose derivative with respect to $x$ is $$H'(x)=\frac{1}{G(x)}\quad\left(\text{solution: $H(x)=\int \frac{dx}{G(x)}$}\right).\tag{2}$$ The chain rule then implies that the left hand side in $(1)$ can be written as $$\frac{1}{G(x(t))}\frac{dx}{dt}=H'(x(t))\frac{dx}{dt}=\frac{dH(x(t))}{dt}.$$ Therefore $(1)$ is equivalent to $$\frac{dH(x(t))}{dt}=F(t).$$ In your example $$F(t)=a,\quad G(x)=x.$$ Therefore $$H'(x)=\frac{1}{x},\quad a=F(t)=\frac{d}{dt}\left(H(x(t))\right)=\frac{d}{dt}\left(\ln(x(t))\right)=\frac{1}{x}\frac{dx}{dt},$$ which implies $$\frac{dx}{x}=a\,dt.$$ It is more intuitive to observe that the first order differential equation $$\frac{dx}{dt}=ax\tag{3}$$ is separable. Thus, it can be rewritten as $$\frac{dx}{x}=a\,dt.\tag{4}$$

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