Question

СРОЧНО
Решите уравнение
(x^2 + 2xy) dx+ xydy=0

Answer 1

Ответ:

$(x^2+2xy)\, dx+xy\, dy=0\\\\\\\displaystyle \frac{dy}{dx}=-\frac{x^2+2xy}{xy}\ \ ,\ \ \ \frac{dy}{dx}=-\frac{x}{y}+2\\\\\\t=\frac{y}{x}\ \ ,\ \ y=tx\ \ ,\ \ y'=t'x+t\\\\\\t'x+t=\frac{1}{t}+2\ \ ,\ \ \ t'x+t=\frac{1+2t}{t}\ \ ,\ \ \ t'x=\frac{1+2t}{t}-t\ \ ,\\\\\\t'x=\frac{1+2t-t^2}{t}\ \ ,\ \ \ t'=-\frac{t^2-2t-1}{tx}\ \ ,\ \ \ \frac{dt}{dx}=-\frac{t^2-2t-1}{tx}\ \ ,\\\\\\\int \frac{t\, dt}{t^2-2t-1}=-\int \frac{dx}{x}$

$\displaystyle \int \frac{t\, dt}{t^2-2t-1}=\int \frac{t\, dt}{(t-1)^2-2}=\Big[\ t-1=z\ ,\ dt=dz\ \Big]=\\\\\\=\int \frac{(z+1)\, dz}{z^2-2}=\frac{1}{2}\int \frac{2z\, dz}{z^2-2}+\int \frac{dz}{z^2-2}=\\\\\\=\frac{1}{2}\, ln|z^2-2|+\frac{1}{2\sqrt2}\cdot ln\Big|\, \frac{z-\sqrt2}{z+\sqrt2}\, \Big|+C=\\\\\\=\frac{1}{2}\, ln\Big|\frac{y^2}{x^2}-2\Big|+\frac{1}{2\sqrt2}\cdot ln\Big|\frac{\frac{y}{x}-\sqrt2}{\frac{y}{x}+\sqrt2} \, \Big|+C=$

$\displaystyle =\frac{1}{2}\, ln\Big|\frac{y^2-2x^2}{x^2}\Big|+\frac{1}{2\sqrt2}\cdot ln\Big|\frac{\frac{y}{x}-\sqrt2}{\frac{y}{x}+\sqrt2} \, \Big|+C=\\\\\\=\frac{1}{2}\, ln\Big|\frac{y^2-2x^2}{x^2}\Big|+\frac{1}{2\sqrt2}\cdot ln\Big|\frac{y-x\sqrt2}{y+x\sqrt2} \, \Big|+C\\\\\\Otvet:\ \frac{1}{2}\, ln\Big|\frac{y^2-2x^2}{x^2}\Big|+\frac{1}{2\sqrt2}\cdot ln\Big|\frac{y-x\sqrt2}{y+x\sqrt2} \, \Big|=-ln|x|-lnC\ \ ,$

${}\qquad \ \ \ \ \ \ \ \dfrac{1}{2}\, ln\Big|\dfrac{y^2-2x^2}{x^2}\Big|+\dfrac{1}{2\sqrt2}\cdot ln\Big|\dfrac{y-x\sqrt2}{y+x\sqrt2} \, \Big|=-ln|Cx|$