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.

  • $\begingroup$ 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

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 $$\frac{dx}{x}=a\,dt.\tag{4}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.