Solutions of first order Ordinary differential equation

Question

I'm given a multiple choice question of the following. But, I find it difficult to think of a way to tackle the problem. Can I have some help here?

Which of the following functions is a solution of some differential equation of the form $y’=f(y)$, where $f$ is continuously differentiable for all real $y$?

a) $y = \sin(t)$

b) $y = t^2$

c) $y = t^3 - t$

d) $y = \cosh(t)$

e) $y = \tanh(t)$

f) all of the above

Edit: after getting help and by looking at the graph of the following functions, I could see that only $y=\tanh(t)$ is possible.

I was also told to write out $y'$ and express it in terms of $y$, but i could not do it for all choices.

Any other methods to tackle such a problem?

Answer 1

On the one hand, this equation is autonomous because the right-hand doesn't explicitly depend on $t$. For autonomous equations, shifts of solutions are also solutions; in other words, if $y=g(t)$ is a solution, so is $y=g(t-c)$ for any $c$.

On the other hand, since $f$ is given to be continuously differentiable, this equations satisfies the conditions for existence and uniqueness of solutions. So there must be precisely one and only one solution thru any point $(t_0,y_0)$.

If you put these two facts together, you'll see that each solution to such an equation has to be a one-to-one function on $\mathbb{R}$. Because if it isn't, then you can shift it and have two solution curves thru the same point. For example, if $y=t^2$ is a solution, so is $y=(t-2)^2$, and we have two solution curves thru $(1,1)$:

The only one-to-one function among the given answer choices is $y=\tanh(t)$.