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When we have a first-order differential equation such as

$$ y'(t) = y(t) $$

the general solution yields

$$ y(t) = c_1 \cdot e^t $$

where $c_1$ is a constant. Notice that the differential equation is autonomous ($y'$ does not depend on t in the equation directly). My question is now the following:

Can any function $y(t)$ be expressed as a first order autonomous differential equation? I am struggeling with the following equations

$y = t^a$ where a is an arbitrary real value and $y=ln(t)$. For the polynomial I tried

$$ y' = a \cdot t^{a-1}$$ $$ y' = a \cdot t^a \cdot t^{-1}$$ $$y' = a \cdot y \cdot \frac{d}{dt}ln(t)$$

but I am unable to express either this equation, nor $y=ln(t)$ itself in such a form, that the equation will only have $y$ or its derivative in it.

Murad Babayev
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  • Do you mean, for every (say, $C^1$) function $y : (a, b) \to \Bbb R$ is there an first order autonomous differential equation, i.e. a function $F : y((a, b)) \to \Bbb R$ such that $y$ is a solution of the autonomous differential equation $z' = F(z)$? – Travis Willse Mar 06 '19 at 19:36
  • Notice, for example, that for $y(t) = \log t$, we can write $y' = 1 / t$ as $1 / \exp \log t = \exp (-\log t)$, so $y(t)$ is a solution of the autonomous equation $y' = \exp(-y)$. – Travis Willse Mar 06 '19 at 19:39
  • No, not any function can be expressed as a solution of an autonomous differential equation: it should be monotone. See, e.g., Proof by contradiction that there are no periodic solutions to $\dot{x}=f(x)$. – user539887 Mar 06 '19 at 19:40
  • @user539887 Thank you, you both actually helped me out. Is there a general rule of what order x has to be so that the equation holds for a given f(x)? – Murad Babayev Mar 06 '19 at 19:52
  • I don't know of any. – user539887 Mar 06 '19 at 20:04
  • @user539887 But isn't $y' = \sqrt{1-y^2}$ periodic? Anyways, if your claim is right, then there should still exist an autonomous diffeq. for $y = t^a$ since it is monotone. – Murad Babayev Mar 07 '19 at 14:02
  • No solution of $y'=\sqrt{1-y^2}$ is periodic. You can easily find, adapting the method given by Travis, an ODE one of whose solutions is $y(t)=t^{\alpha}$. – user539887 Mar 07 '19 at 14:21
  • @user539887 I am getting from multiple sources that $y = sin(c_1 + t)$ is a soultion to this ODE. I must say that I cannot verify this as I do not have the proper knowledge to solve such an equation by myself.

    Regarding my question I found $y' = a \cdot y^{\frac{a-1}{a}}$. I am not sure if this is correct though.

     – Murad Babayev Mar 07 '19 at 14:41
  • You probably use some sources where formal calculations are made. Any (nonextendible) solution to $y'=\sqrt{1-y^2}$ has the form \begin{equation}\begin{cases}-1&\text{for }t\in(-\infty,-c_1-\tfrac{\pi}{2})\ \sin{(t+c_1)}&\text{for }t\in[-c_1-\tfrac{\pi}{2},-c_1+\tfrac{\pi}{2}]\ 1&\text{for }t\in(-c_1+\tfrac{\pi}{2},\infty)\end{cases}\end{equation}(and you can see that the above functions are not periodic), plus the functions equal constantly to $-1$ and to $1$. Your last formula is correct. – user539887 Mar 07 '19 at 18:55

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