Question

Помогите!!! Очень надо!!!

Answer 1

Ответ:

$4)\ \ y'=\dfrac{x+2y}{2y-x}\\\\u=\dfrac{y}{x}\ \ ,\ \ y=ux\ \ ,\ \ y'=u'x+u\\\\u'x+u=\dfrac{x+2ux}{2ux-x}\ \ ,\ \ \ u'x+u=\dfrac{1+2u}{2u-1}\ \ ,\ \ u'x=\dfrac{1+2u}{2u-1}-u\ ,\\\\\\u'x=\dfrac{1+2u-2u^2+u}{2u-1}\ \ ,\ \ \ u'x=\dfrac{-2u^2+3u+1}{2u-1}\ \ ,\ \ \dfrac{du}{dx}=\dfrac{-(2u^2-3u-1)}{x\, (2u-1)}\\\\\\\int \dfrac{(2u-1)\, du}{2u^2-3u-1}=-\int \dfrac{dx}{x}$

$\int \dfrac{(2u-1)\, du}{2u^2-3u-1}=\dfrac{1}{2}\int \dfrac{(2u-1)\, du}{(u-\frac{3}{4})^2-\frac{17}{16}}=\Big[\ t=u-\dfrac{3}{4}\ \Big]=\dfrac{1}{2}\int \dfrac{(t+\frac{1}{2})\, dt}{t^2-\frac{17}{16}}=\\\\\\=\dfrac{1}{4}\int \dfrac{2t\, dt}{t^2-\frac{17}{16}}+\dfrac{1}{4}\int \dfrac{dt}{t^2-\frac{17}{16}}=\dfrac{1}{4}ln\Big|\, t^2-\dfrac{17}{16}\, \Big|+\dfrac{1}{4}\cdot ln\Big|\, \dfrac{t-\frac{\sqrt{17}}{4}}{t+\frac{\sqrt{17}}{4}}\, \Big|+C=$

$=\dfrac{1}{4}ln\Big|\, \Big(\dfrac{y}{x}-\dfrac{3}{4}\Big)^2-\dfrac{17}{16}\, \Big|+\dfrac{1}{4}\cdot ln\Big|\, \dfrac{4(\frac{y}{x}-\frac{3}{4})-\sqrt{17}}{4(\frac{y}{x}-\frac{3}{4})+\sqrt{17}}\, \Big|+C\ ;\\\\\\\dfrac{1}{4}ln\Big|\, \Big(\dfrac{y}{x}-\dfrac{3}{4}\Big)^2-\dfrac{17}{16}\, \Big|+\dfrac{1}{4}\cdot ln\Big|\, \dfrac{4(\frac{y}{x}-\frac{3}{4})-\sqrt{17}}{4(\frac{y}{x}-\frac{3}{4})+\sqrt{17}}\, \Big|+C=-ln|x|+C$

$5)\ \ y'\, tgx-y=1\ \ ,\ \ \ y'-\dfrac{1}{tgx}\cdot y=\dfrac{1}{tgx}\\\\y=uv\ ,\ \ y'=u'v+uv'\\\\u'v+uv'-ctgx\cdot uv=ctgx\\\\u'v+u\, (v'-ctgx\cdot v)=ctgx\\\\a)\ \ \dfrac{dv}{dx}=ctgx\cdot v\ \ ,\ \ \ \int \dfrac{dv}{v}=\int ctgx\cdot dx\ \ ,\ \ ln|v|=ln|sinx|\ \ ,\\\\v=sinx\\\\b)\ \ u'v=ctgx\ \ ,\ \ \dfrac{du}{dx}\cdot sinx=ctgx\ \ ,\ \ \ \int du=\int \dfrac{ctgx}{sinx}\, dx\ \ ,\\\\\int du=\int \dfrac{cosx}{sin^2x}\, dx\ \ ,\ \ \ u=-\dfrac{1}{sinx}+C$

$c)\ \ y=sinx\, \Big(C-\dfrac{1}{sinx}\Big)\ \ ,\ \ \ \ \underline {y=C\, sinx-1\ }$

$6)\ \ y'-\dfrac{y}{x}=\dfrac{1+\sqrt{x}}{x}\\\\y=uv\ \ ,\ \ \ y'=u'v+uv'\\\\u'v+uv'-\dfrac{uv}{x}=\dfrac{1+\sqrt{x}}{x}\ \ ,\ \ \ u'v+u\, \Big(v'-\dfrac{v}{x}\Big)=\dfrac{1+\sqrt{x}}{x}\ \ ,\\\\\\a)\ \ \dfrac{dv}{dx}=\dfrac{v}{x}\ \ ,\ \ \ \int \dfrac{dv}{v}=\int \dfrac{dx}{x}\ \ ,\ \ \ln|v|=ln|x|\ \ ,\ \ v=x\\\\b)\ \ \dfrac{du}{dx}\cdot x=\dfrac{1+\sqrt{x}}{x}\ \ ,\ \ \ \int du=\int \dfrac{1+\sqrt{x}}{x^2}\, dx\ ,$

$u=\int \dfrac{dx}{x^2}+\int \dfrac{dx}{x^{3/2}}\ \ ,\ \ \ u=-\dfrac{1}{x}-\dfrac{3}{2\sqrt{x}}\ \ ,\\\\\\c)\ \ y=x\cdot \Big(-\dfrac{1}{x}-\dfrac{3}{2\sqrt{x}}\Big)\ \ ,\ \ \ \ \underline{\ y=-1-\dfrac{3}{2}\, \sqrt{x}\ }$