I have the function $$\frac{dV}{dT}=1-V^2$$
Just looking to see if my working is okay.
$$dV=1-V^2dT$$ $$\frac{1}{1-V^2}dV=dT$$
Integrate
$$\int{}\frac{1}{1-V^2}dV=\int{}dT$$ Let $V=\tanh(x)$
$\frac{dV}{dx}=sech^2(x)$
$dV=sech^2(x)dx$
$$\int{}\frac{1}{1-\tanh^2(x)}sech^2(x)dx=\int{}dT$$
$1-tanh^2(x)=sech^2(x)$ $$\int{}\frac{1}{sech^2(x)}sech^2(x)dx=\int{}dT$$ $$\int{}\frac{sech^2(x)}{sech^2(x)}dx=\int{}dT$$ $$\int{}\ dx=\int{}\ dT$$ $$x=\int{}dT$$ $V=\tanh(x)$ so take $\tanh^{-1}$ of both sides to find x $$\tanh^{-1}(V)=x$$ $$\tanh^{-1}(V)=\int{}dT$$ $$=T+C$$ Take tanh of both side $$V=\tanh(T+C)$$
Therefore the general solution of $\frac{dV}{dT}=1-V^2$ is $V=\tanh(T+C)$
It is then asking for the particular solution of this for when t=0, v=0 for V is a function of T. I have no idea what they mean.
Rewrite $$ \begin{align} \int{}\frac{1}{1-V^2}dV&=\frac{1}{2}\int{}\left(\frac{1}{1-V}+\frac{1}{1+V}\right)dV\\ &=\frac{1}{2}\left(\int\frac{1}{1-V}\ dV+\int\frac{1}{1+V}\ dV\right)\\ &=-\frac{1}{2}\ln(1-V)+\frac{1}{2}\ln(1+V)+C\\ &=\frac{1}{2}\ln\left(\frac{1+V}{1-V}\right)+C \end{align} $$ then $$ \begin{align} T&=\frac{1}{2}\ln\left(\frac{1+V}{1-V}\right)+C\\ \ln\left(\frac{1+V}{1-V}\right)&=2(T-C)\\ \frac{1+V}{1-V}&=e^{2(T-C)}\\ \frac{1+V}{1-V}&=Ke^{2T}\quad\Rightarrow\quad K=e^{-2C}\\ 1+V&=Ke^{2T}-Ke^{2T}V\\ V+Ke^{2T}V&=Ke^{2T}-1\\ (1+Ke^{2T})V&=Ke^{2T}-1\\ V&=\frac{Ke^{2T}-1}{Ke^{2T}+1}. \end{align} $$ Note that, your solution is also correct.
Now, plugging in the initial condition where $V=0$ when $T=0$. The purpose of the initial condition is to obtain the value of $K$, in your case $C$. \begin{align} 0&=\frac{Ke^{2(0)}-1}{Ke^{2(0)}+1}\\ 0&=\frac{K-1}{K+1}\\ K-1&=0\\ K&=1 \end{align} Thus, $$ V(T)=\frac{e^{2T}-1}{e^{2T}+1}. $$