Qualitative Treatment of Autonomous ODE

A differential equation is called autonomous if it has no independent term. Hence a first order general ode will look like
$$\dfrac{dy}{dt} = f(y) \phantom{csir net}\text{ [Independent of $t$]}.$$

Qualitative Treatment of Autonomous ODE
STEP I: Find the critical points $y_0$(solution of $f(y)=0$). These are horizontal solution of the given ode.

Since two solutions(called integral curve) can’t intersect, hence other solution will be in the region bounded by these horizontal solution.
STEP II: As we have partition the space with horizontal solution, now identify the region with $f(y)>0$ and $f(y)<0$.

As we know that
$$\begin{align*} \left [\dfrac{dy}{dt} = f(y) \right ] \begin{matrix} >0 &\Rightarrow y(t)\text{ is increasing}\\ <0 &\Rightarrow y(t)\text{ is decreasing} \end{matrix} \end{align*}$$
Hence the solution will look like

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Now we will apply these concept to a question of CSIR-NET.

Question: Let $y:\mathbb{R} \rightarrow \mathbb{R}$ satisfy the initial value problem
$$y'(t)=1-y^2,t\in \mathbb{R},~y(0)=0.$$ Then

1. $y(t_1)=1$ for some $t\in R$.
2. $y(t)>-1$ for all $t\in R$.
3. $y$ is strictly increasing in $\mathbb{R}$.
4. $y$ is increasing in $(0, 1)$ and decreasing in $(1,\infty)$.

Answer: Here $f(y)=1-y^2=(1-y)(1+y)$, so $y=\pm 1$ are two horizontal solution of the given ode
$$\begin{matrix} \text{Region 1}&y > 1 & f(y) < 0 & y(t) \text{ is decreasing}\\ \text{Region 2}&-1>y>1 & f(y) > 0 &y(t) \text{ is increasing}\\ \text{Region 3}&y <-1 & f(y) < 0 & y(t) \text{ is decreasing}\\ \end{matrix}$$
The required integral curve will pass through $(0,0)$, which lies in region 2, so the solution is increasing and bounded by $y \in (-1,1)$. Here 2nd and 4th option are correct.

Hence 2 and 4 are correct choices.