Question

x(x+2y)dx+(x^2-y^2)dy=0
помогите пожалуйста

Answer 1

Ответ:

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

$\displaystyle t'x=\frac{t^2-3t-1}{1-t}\ \ ,\ \ \ \frac{dt}{dx}\cdot x=\frac{t^2-3t-1}{1-t}\ \ ,\ \int \frac{(1-t)\, dt}{t^2-3x-1}=\int \frac{dx}{x}\ \ ,\\\\\\\star \ \ \int \frac{(1-t)\, dt}{t^2-3x-1}=\int \frac{(1-t)\, dt}{(t-\frac{3}{2})^2-\frac{9}{4}-1}=\int \frac{(1-t)\, dt}{(t-\frac{3}{2})^2-\frac{13}{4}}=\Big[\ z=t-\frac{3}{2}\ \Big]=\\\\\\=\int \frac{(-z-\frac{1}{2})\, dz}{z^2-\frac{13}{4}}=-\int \frac{z\, dz}{z^2-\frac{13}{4}}-\frac{1}{2}\int \frac{dz}{z^2-\frac{13}{4}}=$

$=-\dfrac{1}{2}\cdot ln\Big|\, z^2-\dfrac{13}{4}\, \Big|-\dfrac{1}{2}\cdot \dfrac{1}{2\cdot \frac{\sqrt{13}}{2}}\cdot ln\Big|\, \dfrac{z-\frac{\sqrt{13}}{2}}{z+\frac{\sqrt{13}}{2}}\, \Big|+C=\\\\\\=-\dfrac{1}{2}\cdot ln\Big|\, z^2-\dfrac{13}{4}\, \Big|-\dfrac{1}{2\cdot \sqrt{13}}\cdot ln\Big|\, \dfrac{2z-\sqrt{13}}{2z+\sqrt{13}}\, \Big|+C=\\\\\\=-\dfrac{1}{2}\cdot ln\Big|\, (t-\frac{3}{2})^2-\dfrac{13}{4}\, \Big|-\dfrac{1}{2\cdot \sqrt{13}}\cdot ln\Big|\, \dfrac{2t-3-\sqrt{13}}{2t-3+\sqrt{13}}\, \Big|+C=$

$=-\dfrac{1}{2}\cdot ln\Big|\, \Big(\dfrac{y}{x}-\dfrac{3}{2}\Big)^2-\dfrac{13}{4}\, \Big|-\dfrac{1}{2\cdot \sqrt{13}}\cdot ln\Big|\, \dfrac{\frac{2y}{x}-3-\sqrt{13}}{\frac{2y}{x}-3+\sqrt{13}}\, \Big|+C=\\\\\\=-\dfrac{1}{2}\cdot ln\Big|\, \dfrac{y^2}{x^2}-\dfrac{3y}{x}-1\, \Big|-\dfrac{1}{2\cdot \sqrt{13}}\cdot ln\Big|\, \dfrac{2y-3x-\sqrt{13}\, x}{2y-3x+\sqrt{13}\, x}\, \Big|+C\ \ \star$

$Otvet:\\\\-\dfrac{1}{2}\cdot ln\Big|\, \dfrac{y^2}{x^2}-\dfrac{3y}{x}-1\, \Big|-\dfrac{1}{2\cdot \sqrt{13}}\cdot ln\Big|\, \dfrac{2y-3x-\sqrt{13}\, x}{2y-3x+\sqrt{13}\, x}\, \Big|+C=ln|x|+C$