(y−xy2)dx+(x+x2y2)dy=0;when y(1)=1

Solution:
${\displaystyle (4-x{y}^2)dx+(x+x^2{y}^2)dx=0}$
${\displaystyle (4dx+xdy)-(x{y}^{2}dy-x^2{y}^2)=0}$
${\displaystyle d\left(xy\right)-{\left(xy\right)}^2[\frac{dx}{x}-dy]=0}$
${\displaystyle \int \frac{d\left(xy\right)}{{\left(xy\right)}^2}-\int \frac{dx}{x}+\int dy=0}$
${\displaystyle \Rightarrow -\frac{1}{x\cdot y}-\ln x+y=C}$
put ${\displaystyle 4\left(1\right)=1}$
${\displaystyle -\frac{1}{1\cdot 1}-\ln 1+1=C\Rightarrow -1-0+1=C}$
${\displaystyle c=0}$
Hence, ${\displaystyle -\frac{1}{x\cdot y}\cdot \ln x+y=0}$

Consider the differential equation
${\displaystyle (y-x{y}^2)dx+(x+x^2{y}^2)dy=0}$
${\displaystyle ydx-x{y}^{2}dx+xdy+x^2{y}^{2}dy=0}$
${\displaystyle (ydx+xdy)+(x^2{y}^{2}dy-x{y}^{2}dx)=0}$
${\displaystyle d\left(xy\right)+xy(xydy-ydx)=0}$
${\displaystyle d\left(xy\right)+x^2{y}^{2}dy-x{y}^{2}dx=0}$
Dividing by ${\displaystyle x^2{y}^2}$
${\displaystyle d\frac{xy}{x^2{y}^2}+dy-\frac{dx}{x}=0}$
${\displaystyle d\frac{xy}{{\left(xy\right)}^2}+dy-\frac{dx}{x}=0}$
Integrating the equation above we get
${\displaystyle \int \left[d\frac{xy}{{\left(xy\right)}^2}\right]+\int dy-\int \frac{dx}{x}=C}$…..(1)
Where C is constant of integration
Now say ${\displaystyle \int \frac{dz}{z^2}=-\frac{1}{z}}$
${\displaystyle \int x^{n}dx=x^n+\frac{1}{n+1}+c\text{for}n\ne -1}$
${\displaystyle \int \frac{dx}{x}=\ln x+c}$
Using these identities if ${\displaystyle z=xy}$ then the above eqn.(1) becomes
${\displaystyle -\frac{1}{xy}+y-\ln x=C}$
Multiply throughout by xy in the above equation we get
${\displaystyle -1+x{y}^2-xy\ln x=Cxy}$
Therefore the solution of the equation is
${\displaystyle x{y}^2-xy\ln x-1=Cxy}$.
Where C is any arbitrary constant.