Question

Дифференциальные уравнения
1) y'x+x+y=0 2) y'+2xy=2xy^3

Answer 1

$1)\ \ y'x+x+y=0\ |:x\ \ ,\ \ \ \ y'=-\dfrac{y}{x}-1\\\\u=\dfrac{y}{x}\ \ ,\ \ y=ux\ \ ,\ \ y'=u'x+u\\\\u'x+u=-u-1\ \ ,\ \ u'x=-2u-1\ \ ,\ \ u'=-\dfrac{2u+1}{x}\ ,\\\\\int \dfrac{du}{2u+1}=-\int \dfrac{dx}{x}\\\\\dfrac{1}{2}\, ln|2u+1|=-ln|x|+lnC\ \ ,\ \ \ \sqrt{2u+1}=\dfrac{C}{x}\ \ ,\\\\\sqrt{\dfrac{2y}{x}+1}=\dfrac{C}{x}\ \ ,\ \ \dfrac{2y+x}{x}=\dfrac{C^2}{x^2}\ \ ,\ \ 2y+x=\dfrac{C^2}{x}\ \ ,\ \ y=\dfrac{C^2}{2x}-\dfrac{x}{2}\\\\\\y=\dfrac{C^*}{x}-\dfrac{x}{2}\ \ \Big(\ C^*=\dfrac{C^2}{2}\ \Big)$

$2)\ \ y'+2xy=2xy^3\ \ ,\ \ \ \ y=uv\ ,\ \ y'=u'v+uv'\\\\u'v+uv'+2xuv=2x(uv)^3\\\\u'v+u(v'+2xv)=2x(uv)^3\\\\a)\ \ \dfrac{dv}{dx}=-2xv\ \ ,\ \ \ \int \dfrac{dv}{v}=-2\int x\, dx\ \ ,\ \ ln|v|=-x^2\ \ ,\ \ v=e^{-x^2}\\\\b)\ \ \dfrac{du}{dx}\cdot e^{-x^2}=2x\cdot u^3\cdot e^{-3x^2}\\\\\int \dfrac{du}{u^3}=\int 2x\cdot e^{-2x^2}\, dx\ \ ,\ \ \ \dfrac{u^{-2}}{-2}=-\dfrac{1}{2}\int e^{-2x^2}\cdot d(-2x^2)\ ,\\\\-\dfrac{1}{2u^2}=-\dfrac{1}{2}e^{-2x^2}-\dfrac{C}{2}\ \ ,\ \ \ \dfrac{1}{u^2}=e^{-2x^2}+C\ \ ,$

$u=\pm \dfrac{1}{\sqrt{e^{-2x^2}+C}}\ \ \ \Rightarrow \ \ \ \ y=\pm \dfrac{e^{-x^2}}{\sqrt{e^{-2x^2}+C}}$