# Picard-Lindelöf Theorem application

I'm really lost here guys. I'd appreciate it if you could help.

Consider the differential equation

$$\frac{dy}{dt} = f(t, y) \tag{1}$$

with $$f$$ satisfying the conditions of the Picard-Lindelöf Theorem. Also, $$y_1(t) = 3\ ,\ t \in \mathbb{R}$$ is a solution of $$(1)$$. What can we conclude about the solution $$y(t)$$ which satisfies the initial condition $$y(0) = 1$$ ?

So, for anyone interested, after a couple of days and with some indications, I solved my question.

Let $y_2(t)$ be the solution with initial condition $y(0) = 1$.

Now suppose that $y_1$ and $y_2$ have a common point at $(t_0, y_1(t_0)) =(t_0, 1)$. Then the following problem

$$\frac{dy}{dt} = f(t, y)\ ,\ y(t_0) = 1$$

has 2 solutions, $y_1$ and $y_2$. But that can't be true because of the Picard-Lindelof Theorem. So $y_1(t) \neq y_2(t)$ for all $t \in \mathbb{R}$ and because $y_2(0) < y_1(t)$ we conclude that $y_2(t) < y_1(t) \implies y_2(t) < 3$ for all $t \in \mathbb{R}$.