The solution of the differential equation $\mathrm{ydx}-(\mathrm{x}+2{\mathrm{y}}^2)\mathrm{dy}=0$ is $x=\text{⨍ (y).}$ If $\text{⨍ (-1)=1,}$ then $\text{⨍ (1)}$ is equal to :

$\begin{array}{l}ydx=(x+2y^2)dy\\ \Rightarrow \frac{dx}{dy}=\frac{x}{5}+2y\\ \Rightarrow \frac{dx}{dy}-\frac{1}{y}.x=2y\\ I.F=e^{\int -\frac{1}{y}dy}=e^{-\log y}=e^{\log \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$y$}\right.\\ Solutionis\\ x(I.F)=\int 2y(I.F)dy\\ =2\int y\left(\frac{1}{y}\right)dy\\ \frac{x}{y}=2y+c\end{array}$

$\begin{array}{l}\frac{f\left(y\right)}{y}=2y+c\\ Puty=-1\\ \frac{f(-1)}{-1}=-2+c\\ \Rightarrow -1=-2+c\\ \Rightarrow c=1\\ \frac{f\left(y\right)}{y}=2y+1\\ Puty=1\\ \frac{f\left(1\right)}{1}=2\left(1\right)+1\\ f\left(1\right)=3\end{array}$