How to rewrite $\frac{dx}{dt}=a\cdot x$ into the form $\frac{dx}{x}=a\cdot dt$ using the chain rule

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.

• Perhaps they are thinking of $$dt = \frac{dt}{dx} dx$$ as an expression of the chain rule. – WW1 Dec 19 '20 at 5:09
• @WW1 yes I think that's what they meant if: given $\frac{dx}{dt}$ then $dx=\frac{dx}{dt}dt$. – Alexander Orman Dec 19 '20 at 5:36

$$\frac{dx}{dt}=F(t)G(x).$$ 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 $$\frac{dx}{x}=a\,dt.\tag{4}$$