Ответ:

[tex]1)\ \ (1+x)\, y\, dx+(1-y)\, x\, dy=0[/tex]

Дифф. уравнение 1 пор. с разделяющимися переменными .

[tex]\displaystyle \int \dfrac{(1+x)\, dx}{x}=-\int \frac{(1-y)\, dy}{y}\\\\\\\int \Big(\frac{1}{x}+1\Big)\, dx=-\int \Big(\frac{1}{y}-1\Big)\, dy\\\\\\\underline {ln|x|+x=-ln|y|+y+C\ }\\\\\\2)\ \ (x+y)\, dx+x\, dy=0[/tex]  

Дифф. ур-е 1 пор. с однородными функциями .

[tex]\displaystyle \frac{dy}{dx}=-\frac{x+y}{x}\ \ ,\ \ y'=-\frac{x+y}{x}\ \ ,\ \ \ y'=-1-\frac{y}{x} \ \\\\\\Zamena:\ \ t=\frac{y}{x}\ ,\ y=tx\ ,\ y'=t'x+t\\\\t'x+t=-1-t\ \ ,\ \ t'x=-1-2t\ \ ,\ \ \frac{dt}{dx}=-\frac{1+2t}{x}\ \ ,\\\\\\\int \frac{dt}{1+2t}=-\int \frac{dx}{x}\ \ ,\\\\\\\frac{1}{2}\cdot ln|1+2t|=-ln|x|-lnC\ \ \ \to \ \ \ \sqrt{1+2t}=\frac{1}{Cx}\\\\\\\sqrt{1+\frac{2y}{x}}=\frac{1}{Cx}\ \ ,\ \ \ \sqrt{\frac{x+2y}{x}}=\frac{1}{Cx}\ \ ,\ \ \underline{\ C\sqrt{x\, (x+2y)}=1\ }[/tex]

[tex]\displaystyle 2)\ \ x^2\cdot \frac{dy}{dx}=2xy+3\ \ \to \ \ \ x^2\cdot y'=2xy+3\ \ \to \\\\\\y'-\frac{2}{x}\cdot y=\frac{3}{x^2}[/tex]

Линейное дифф. ур-е 1 порядка .  Замена:  [tex]y=uv\ ,\ y'=u'v+yv'[/tex]  .

[tex]\displaystyle u'v+uv'-\frac{2}{x}uv=\frac{3}{x^2}\ \ ,\ \ \ u'v+u\cdot \Big(v'-\frac{2}{x}\, v\Big)=\frac{3}{x^2}\\\\a)\ \ v'-\frac{2}{x}\, v=0\ \ ,\ \ \int \frac{dv}{v}=\int \frac{2\, dx}{x}\ \ ,\ \ ln|v|=2\, ln|x|\ \ ,\ \ \underline {v=x^2}\ .[/tex]

[tex]\displaystyle b)\ \ u'v=\frac{3}{x^2}\ \ ,\ \ \frac{du}{dx}\cdot x^2=\frac{3}{x^2} \ \ ,\ \ \int du=3\int \frac{dx}{x^4}\ \ ,\ \ u=3\cdot \frac{x^{-3}}{-3}+C\ ,\\\\\\\underline{\ u=-\frac{1}{x^3}+C\ }\\\\\\c)\ \ y=x^2\, \Big(-\frac{1}{x^3}+C\Big)\ \ ,\ \ \ \underline{\ y=-\frac{1}{x}+Cx^2\ }[/tex]