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Let consider the system: $$\begin{cases}\dot x=x\\ \dot y=-y+x^2.\end{cases}$$

To solve this system, my teacher made: $$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=(-y+x^2)\frac{1}{dx/dt}=(-y+x^2)\frac{1}{x}.$$

And thus $y(x)=\frac{x^2}{3}+\frac{c}{x}.$

Question: Does this method always work? And if yes why?

Because I have the impression that it really simplify the problem, and I know that: $$\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dx}$$ is not always true as this can show.
So why does it work here ?

amWhy
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user621345
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  • Note that the solution is not complete at this point, as the original independent variable is $t$. But here you can trivially solve the first equation and then insert into the found formula. – Lutz Lehmann Dec 18 '18 at 13:59
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    You linked to a question about partial derivatives. Your system uses ordinary derivatives. – David K Dec 18 '18 at 14:27
  • There are involved two equivalent (?) representation kinds. A parametric one $(x(t), y(t))$ and an explicit one after $t$ elimination $y = f(x)$ – Cesareo Dec 18 '18 at 14:38
  • @Cesareo Not quite equivalent. The solutions $F(x,y) = c$ are a family of curves which fill the phase plane, called the orbits. We can determine all points which are reached on a solution $(x(t),y(t))$ for all $t$, given $(x(0),y(0))$. This is enough if $t$ is just a parameter and nothing else, but often, $t$ is a physical variable (such as time) and we may want to know the solution at a particular time $t^*$ or what happens as $t\to\infty$, which cannot be deduced from the $F(x,y)=c$ equation alone. – Alex Jones Dec 19 '18 at 07:34

2 Answers2

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It does always work, and it is called Chain Rule

william122
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If $y(x) = y(x(t))$, then by the chain rule

$$ \frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} $$

Divide out $\frac{dx}{dt}$ from both sides.

Dylan
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