i need more understandable solution of the problem $1$

$(1)(x+y{)}^{2}dx+(2xy+x^2-1)dy=0$

$15$

or $(x^2+2xy+y^2)dx+(2xy+x^2-1)dy=0,\frac{\text{d}\text{e}\text{l}\text{M}}{\text{d}\text{e}\text{l}\text{y}}=\frac{\text{d}\text{e}\text{l}\text{N}}{\text{d}\text{e}\text{l}\text{x}}=>$ exact

$\frac{\text{d}\text{e}\text{l}\text{F}}{\text{d}\text{e}\text{l}\text{x}}=$ and $\frac{\text{d}\text{e}\text{l}\text{F}}{\text{d}\text{e}\text{l}\text{y}}=N$

$=>F={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}dx=\frac{1}{3}(x+y{)}^3+C(y)$

$=>\frac{\text{d}\text{e}\text{l}\text{F}}{\text{d}\text{e}\text{l}\text{y}}=N=>\frac{3}{3}(x+y{)}^2+{}^{\text{'}}(y)=2xy+x^2-1$

$=>x^2+2xy+y^2+{}^{\text{'}}(y)=2xy+x^2-1$

${}^{\text{'}}(y)=-1-y^2$

$(y)=-y-\frac{y^3}{3}+C$

$F(x,y)=\frac{1}{3}(x+y{)}^3-y-\frac{y^3}{3}+C$

or $\frac{1}{3}(x^3+3x^{2}y+3x{y}^2+y^3)-y-\frac{y^3}{3}+c=0$

$\frac{x^3}{3}+x^{2}y+x{y}^2-y=c$

from $y(y)=2$ we get.

$\frac{3^3}{3}+3^2*2+3y{|}_2^2-2||=c=>c=37$

Problem $1.$ Solve the following initial value problem.

$(15$ pts$.)$

$(x+y{)}^{2}dx+(2xy+x^2-1)dy=0,y(3)=2$