Question

Найти общий интеграл дифференциального уравнения


$xy' = \frac{3y^{3}+6yx^{2}}{2y^{2}+ 3x^{2}}$

Answer 1

$xy'=\frac{3y^3+6yx^2}{2y^2+3x^2}\\\\y'=\frac{3y^3+6yx^2}{2xy^2+3x^3}\; \Big |\frac{:x^3}{:x^3}\\\\y'= \frac{3\cdot (\frac{y}{x})^3+6\cdot \frac{y}{x}}{2\cdot (\frac{y}{x})^2+3}\\\\t=\frac{y}{x}\; ,\; \; y=tx\; ,\; \; y'=t'x+tx'=t'x+t\\\\t'x+t=\frac{3t^3+6t}{2t^2+3}\; ,\; \; t'x=\frac{3t^3+6t}{2t^2+3}-t\; ,\; \; t'x=\frac{3t^3+6t-2t^3-3t}{2t^2+3}\; ,\\\\t'x=\frac{t^3+3t}{2t^2+3}\; ,\; \; t'=\frac{dt}{dx}=\frac{t^3+3t}{x(2t^2+3)}\; ,\\\\\int \frac{(2t^2+3)\, dt}{t^3+3t}=\int \frac{dx}{x}\; ,$

$\int \frac{(2t^2+3)\, dt}{t\cdot (t^2+3)}=\int \frac{dt}{t}+\int \frac{t\, dt}{t^2+3}=ln|t|+\frac{1}{2}\int \frac{d(t^2+3)}{t^2+3}=$

$=ln|t|+\frac{1}{2}\cdot ln|t^2+3|+lnC=ln|\frac{y}{x}|+ln\sqrt{\frac{y^2}{x^2}+3}+lnC\; ;\\\\\\ln|\frac{y}{x}|+ln\sqrt{\frac{y^2}{x^2}+3}+lnC=ln|x|\\\\\frac{Cy}{x}\cdot \sqrt {\frac{y^2}{x^2}+3}=x\; \; \; \; obshij\; integral$

$\star \; \; \frac{2t^2+3}{t\, (t^2+3)}=\frac{A}{t}+\frac{Bt+C}{t^2+3}=\frac{A(t^2+3)+t\cdot (Bt+C)}{t\, (t^2+3)}\\\\2t^2+3=A(t^2+3)+t\cdot (Bt+C)\\\\t=0:\; \; A=\frac{3}{3}=1\; ,\\\\2t^2+3=At^2+3A+Bt^2+Ct\\\\t^2\; |\; 2=A+B\; ,\; \; B=2-A=2-1=1\\\\t\; \; |\; 0=C$