Помогите с заданиями, пожалуйста. 1. Найти общее решение дифференциального уравнения с...

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$1)\ \ y'(1+y)=xy\, sinx\\\\\int \dfrac{(1+y)dy}{y}=\int x\cdot sinx\, dx\qquad \Big[\ u=x\ ,\ du=dx,\ dv=sinx\, dx\ ,\ v=-cosx\ \Big]\\\\\int \Big(\dfrac{1}{y}+1\Big)\, dy=-x\cdot cosx+\int cosx\, dx\\\\\\ln|y|+y=-x\cdot cosx+sinx+C\\\\\\2)\ \ y\, dx+(x+y)\, dy=0\\\\\\\dfrac{dy}{dx}=-\dfrac{y}{x+y}\ \ ,\ \ \ y'=-\dfrac{\frac{y}{x}}{1+\frac{y}{x}}\ \ \ ,\ \ \ t=\dfrac{y}{x}\ \ ,\ \ y=tx\ ,\ y'=t'x+t\\\\\\t'x+t=-\dfrac{t}{1+t}\ \ ,\ \ \ t'x=-t-\dfrac{t}{1+t}\ \ ,\ \ \ t'x=\dfrac{-2t-t^2}{1+t}\ \ ,$

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

$\dfrac{y^2}{x^2}+\dfrac{2y}{x}=\dfrac{C^*}{x^2}\ ,\ \ C^*=C^2$

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