To solve the non linear IVP $y'(t)=1-y^2(t)$ 
Let $y: \mathbb{R} \rightarrow \mathbb{R}$ satisfy the initial value problem $$y'(t)=1-y^2(t), \ t \in \mathbb{R},$$ $$y(0)=0$$
  Then which of the following is true
$(A)$ $y(t_1)=1$ for some $t_1 \in \mathbb{R}$
$(B)$ $y(t)>-1$ for all $t \in \mathbb{R}$
$(C)$ $y$ is strictly increasing in $\mathbb{R}$
$(D)$ $y$ is increasing in $(0,1)$ and decreasing in $(1,\infty)$

I am a biology researcher learning differential equations for related work. I did  an undergraduate course in differential equations but have forgotten most of it. Any help would be appreciated. 
 A: You have:
$$\frac{\text{d}y}{\text{d}t}=1-y^2$$
So to solve the IVP you solve instead the equation
$$\int_{y(0)}^y \frac{1}{1-s^2} \text{d}s = \int_0^t 1 \text{d}k.$$
To be a bit sloppy you divided by $1-y^2$ and multiplied by $\text{d}t$ and then integrated over the according domain starting from the respective initial values. That means for $y$ from $y(0)=0$ and for $t$ from $0$. Now you solve this equation for $y$ and have your solution. After this you check which of the answers are true.
This method is called separation of variables. For further information you can look here https://en.wikipedia.org/wiki/Separation_of_variables for example.
A: Using variable separable method, we have
$$ \frac{dy}{1-y(t)^{2}} = dt...(1)$$
Solving for this equation, we get,
$$\frac{1}{2} ln|\frac{1+y(t)}{1-y(t)}| = t + c...(2)$$
where $c$ is some constant. Now we have $y(0)=0$ so substituting in $(2)$ we have $c=0$. Our solution is now $$ y(t)= \frac{e^{2t}-1}{e^{2t}+1}...(3)$$
We now analyse every case.
CASE $1$: $y(t)$ can never be equal to $1$ as then from $(3) $ we will get $-1=1$ which is not possible.
CASE $2$: $y(t)$ is always greater than $-1$ as is evident from $(2)$.  
We know find $y'(t)$ to conclude for cases $(3)$ and $(4)$.
$$ y(t)= \frac{e^{2t}-1}{e^{2t}+1}$$ so we have
$$ y'(t) = \frac{4e^{2t}}{(e^{2t}+1)^{2}}...(4)$$
It is very evident from $(4)$ that y is increasing for all $\mathbb R$ as $y'(t) \geq 0$ for all $\mathbb R$. Hope it helps.
A: By seperation of variables
\begin{align}
& \frac{dy}{dx}  = 1-y^2\\
\implies & \frac{dy}{1-y^2}=dx\\
\implies & \frac{1}{2}\ln \left(\frac{y-1}{y+1}\right) +c =x\\
& \mbox{Using $y(0)=0$ we get $c=0$. Next solve the expression for $y$ }\\
\implies & y(x)=\frac{e^{2x}-1}{e^{2x}+1}\\
\end{align}
One can easily show that the derivative of $y$ is positive thought, therefore  it is increasing, moreover $y \to-1$as $x \to -\infty$ and $y \to 1$as $x \to \infty$. It can be easily shown that $A$ is not true. Hence $B$ and $C$ are the only correct options.}
