Предмет: Математика, автор: nnnsssttt2

Дифференциальное уравнение!

(x+2y)dx-(2x+y)dy=0

Ответы

Автор ответа: NNNLLL54

Ответ:

$(x+2y)\, dx-(2x+y)\, dy=0\\\\(x+2y)\, dx=(2x+y)\, dy\ \ ,\ \ \ \dfrac{dy}{dx}=\dfrac{x+2y}{2x+y}\ \ ,\ \ y'=\dfrac{x+2y}{2x+y}\ \ ,\\\\\\u=\dfrac{y}{x}\ \ ,\ \ y=ux\ \ ,\ \ y'=u'x+u\\\\\\u'x+u=\dfrac{x+2ux}{2x+ux}\ \ ,\ \ \ u'x=\dfrac{1+2u}{2+u}-u\ \ ,\ \ \ u'x=\dfrac{1+2u-2u-u^2}{2+u}\ ,\\\\\\u'x=\dfrac{1-u^2}{2+u}\ \ ,\ \ \ \dfrac{du}{dx}=\dfrac{1-u^2}{x\, (2+u)}\ \ ,\ \ \ \int \dfrac{(2+u)\, du}{1-u^2}=\dfrac{dx}{x}\ \ ,$

$\int \dfrac{2\, du}{1-u^2}-\dfrac{1}{2}\int \dfrac{-2u}{1-u^2}=\dfrac{dx}{x}\ \ ,\\\\\\2\cdot \dfrac{1}{2}\cdot ln\Big|\, \dfrac{1+u}{1-u}\, \Big|-\dfrac{1}{2}\cdot ln|1-u^2|=ln|x|+lnC\\\\\\ln\Big|\, \dfrac{1+\frac{y}{x}}{1-\frac{y}{x}}}\, \Big|-\dfrac{1}{2}\cdot ln\Big|\, 1-\dfrac{y^2}{x^2}\, \Big| =ln|x|+lnC\\\\\\ln\Big|\, \dfrac{x+y}{x-y}\, \Big|-\dfrac{1}{2}\cdot ln\Big|\, \dfrac{x^2-y^2}{x^2}\, \Big|=ln(C|x|)$

$\dfrac{(x+y)\cdot \sqrt{x^2}}{(x-y)\sqrt{x^2-y^2}}=Cx\ \ ,\ \ \ \dfrac{(x+y)\cdot x}{(x-y)\sqrt{(x-y)(x+y)}}=Cx\ \ ,\\\\\\\dfrac{\sqrt{x+y}\cdot x}{x-y}=Cx\ \ \ \to \ \ \ \ \ \boxed{\ \sqrt{x+y}=C(x-y)\ }$