Question

помогите пожалуйста.​

Answer 1

Ответ:

$\displaystyle 1)\ \ y'+\dfrac{y}{x}=xy^2\\\\y=uv\ \ ,\ \ y'=u'v+uv'\\\\u'v+uv'+\dfrac{uv}{x}=x\cdot u^2v^2\ \ ,\ \ \ u'v+u\, (v'+\dfrac{v}{x})=xu^2v^2\ \ ,\\\\a)\ \ v'+\dfrac{v}{x}=0\ \ ,\ \ \int \dfrac{dv}{v}=-\int \frac{dx}{x}\ \ ,\ \ ln|v|=-ln|x|\ \ ,\ \ v=\frac{1}{x}\\\\b)\ \ u'\cdot \frac{1}{x}=x\cdot u^2\cdot \frac{1}{x^2}\ \ ,\ \ \ \frac{du}{dx}=u^2\ \ ,\ \ \int \frac{du}{u^2}=\int dx\ \ ,\\\\-\frac{1}{u}=x+C\ \ ,\ \ u=-\frac{1}{x+C}\\\\c)\ \ y=-\frac{1}{x(x+C)}$

$2)\ \ y''+6y=0\\\\k^2+6=0\ \ ,\ \ k^2=-6\ \ ,\ \ k_{1,2}=\pm \sqrt6\, i\\\\y_{obshee}=C_1\, cos(\sqrt6x)+C_2\, sin(\sqrt6x)$