Ответ:

Диффер. уравнение 1 пор. с однородными функциями .

[tex]\displaystyle \bf (x^2+xy)dx=(y^2+xy)\, dy\\\\\frac{dy}{dx}=\frac{x^2+xy}{y^2+xy}\ \ ,\ \ \ y'=\frac{x^2+xy}{y^2+xy}\ \ \ ,\ \ \ y'=\frac{1+\dfrac{y}{x}}{\dfrac{y^2}{x^2}+\dfrac{y}{x}}[/tex]  

Замена:  [tex]\bf u=\dfrac{y}{x}\ \ ,\ \ \ y=ux\ \ ,\ \ y'=u'x+u[/tex]  .

[tex]\bf \displaystyle u'x+u=\frac{1+u}{u^2+u}\ \ ,\ \ \ u'x+u=\frac{1+u}{u\, (u+1)}\ \ ,\ \ \ u'x+u=\frac{1}{u}\\\\\\u'x=\frac{1}{u}-u\ \ ,\ \ \ u'x=\frac{1-u^2}{u}\ \ ,\ \ \ \frac{x\, du}{dx}=\frac{1-u^2}{u}\ \ ,\\\\\\\int \frac{u\, du}{1-u^2}=\int \frac{dx}{x}\ \ ,\ \ \ -\frac{1}{2}\int \frac{-2u\, du}{1-u^2}=\int \frac{dx}{x}\ \ ,\\\\\\-\frac{1}{2}\cdot ln\Big|1-u^2\, \Big|=ln|\, x\, |+lnC\\\\\\\frac{1}{\sqrt{1-u^2}}=Cx\ \ ,\ \ \ \sqrt{1-u^2}=\frac{1}{Cx}\ \ ,\ \ \ \sqrt{1-\dfrac{y^2}{x^2}}=\frac{1}{Cx}\ ,[/tex]

[tex]\bf \displaystyle \sqrt{\dfrac{x^2-y^2}{x^2}}=\frac{1}{Cx}\ \ ,\ \ \ \frac{x^2-y^2}{x^2}=\frac{1}{C^2x^2}\ \ ,\ \ \ x^2-y^2=\frac{1}{C^2}\ \ ,\\\\\\y^2=x^2-\frac{1}{C^2}\ \ ,\ \ \ y^2=x^2-C_1\ \ ,\ \ \boxed{\ \bf y=\pm \sqrt{x^2-C_1}\ \ ,\ \ \ \Big(C_1=\frac{1}{C^2}\Big)\ }[/tex]