Question

Мат анализ , срочно
100 баллов

Answer 1

Ответ:

решение на фотографии

Answer 2

$(4-y^2x^{-2})\, dx+2yx^{-1}\, dy=0\\\\\\(4-\dfrac{y^2}{x^2})=-\dfrac{2y}{x}\cdot \dfrac{dy}{dx}\ \ ,\\\\\\t=\dfrac{y}{x}\ \ ,\ \ y=tx\ \ ,\ \ y'=t'x+t\ \ ,\ \ 4-t^2=-2t(t'x+t)\\\\\\4-t^2=-2t\, t'\, x-2t^2\ \ ,\ \ 2t\, t'\, x=t^2-4-2t^2\ \ ,\ \ 2t\, t'x=-(t^2+4)\ \ ,\\\\\\t'=-\dfrac{t^2+4}{2t\, x}\ \ ,\ \ \dfrac{dt}{dx}=-\dfrac{t^2+4}{2t\, x}\ \ ,\ \ \int \dfrac{2t\, dt}{t^2+4}=-\int \dfrac{dx}{x}\ \ ,\\\\\\ln|t^2+4|=-ln|x|-lnC\ \ ,\ \ t^2+4=\dfrac{1}{Cx}\ \ ,\ \ t^2=\dfrac{1}{Cx}-4\ \ ,$

$\dfrac{y^2}{x^2}=\dfrac{1}{Cx}-4\ \ ,\ \ y^2=x^2\cdot \dfrac{1-4Cx}{Cx}\ \ ,\ \ \underline {\ y^2=\dfrac{1}{C}\cdot x\cdot (1-4Cx)\ }$