Алгебра: Математика. Дифференциальные уравнения

Математика. Дифференциальные уравнения

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Ответ:

Pikasso07

06.04.2021 06:47

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

$ln|u|=ln|x|+ln|C|\ ,\ \ \ u=Cx\ \ ,\ \ \dfrac{y}{x}=Cx\ \ ,\ \ y_{obshee}=Cx^2\ \ ,\\\\y(1)=1:\ \ 1=C\ \ \ \Rightarrow \ \ \ \underline {\ y_{chastnoe}=x^2\ }$

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

$u=x+C\\\\c)\ \ y_{obshee}=(x+1)^2(x+C)\\\\\\3)\ \ x\, dx+y\sqrt{1+x^2}\, dy=0\ \ ,\ \ \ x\, dx=-y\sqrt{1+x^2}\, dy\ \ ,\\\\\\\int \dfrac{x\, dx}{\sqrt{1+x^2}}=-\int y\, dy\ \ ,\ \ \ \dfrac{1}{2}\cdot 2\sqrt{1+x^2}=-\dfrac{y^2}{2}+C\ \ ,\\\\\\2\sqrt{1+x^2}=-y^2+2C\ \ ,\ \ \ y^2=2C-2\sqrt{1+x^2}\ \ ,\\\\\\y_{obshee}=\pm \sqrt{2C-2\sqrt{1+x^2}}$