Question

Помогите пожалуйста решить

Answer 1

$1)\; \; y'\cdot tgx=1+y\; ,\; \; y( \frac{\pi }{6} )=-\frac{1}{2}\\\\ \frac{dy}{dx} = \frac{1+y}{tgx} \; \; \to \; \; \; \int \frac{dy}{1+y} =\int \frac{dx}{tgx} \\\\\int \frac{d(1+y)}{1+y} =\int \frac{cosx\, dx}{sinx} \\\\\int \frac{d(1+y)}{1+y}=\int \frac{d(sinx)}{sinx} \\\\ln|1+y|=ln|sinx|+ln|C|\\\\1+y=C\cdot sinx\\\\y( \frac{\pi }{6} )=- \frac{1}{2}\; \; \to \; \; 1- \frac{1}{2}=C\cdot sin \frac{\pi}{6} \\\\ \frac{1}{2} =C\cdot \frac{1}{2} \; \; \to \; \; C=1\\\\1+y=sinx\\\\\underline {y=sinx-1}$

$2)\; \; \frac{dy}{\sqrt{x}} +dx= \frac{dx}{\sqrt{y}} \\\\ \frac{dy}{\sqrt{x}}=(\frac{1}{\sqrt{y}} -1)dx\\\\ \frac{dy}{\sqrt{x}} = \frac{1-\sqrt{y}}{\sqrt{y}} dx\\\\\int \frac{\sqrt{y}\, dy}{1-\sqrt{y}} =\int \sqrt{x}\, dx\\\\\int \frac{\sqrt{y}\, dy}{-(\sqrt{y}-1)} =[\; t=\sqrt{y}\; ,\; t^2=y\; ,\; dy=2t\, dt\; ]=-\int \frac{t^2\, dt}{t-1} =\\\\=-\int (t+1+\frac{1}{t-1})dt=- \frac{t^2}{2}-t-ln|t-1|+C_1=\\\\=-\frac{y}{2}-\sqrt{y}-ln|\sqrt{y}-1|+C_1\\\\\int \sqrt{x}\, dx= \frac{2x^{3/2}}{3}+C_2$

$\frac{y}{2}+\sqrt{y}+ln|\sqty{y}-1|+ \frac{2\sqrt{x^3}}{3}+C=0$

Answer 2

$y'tgx=1+y\\y'tgx-y=1\\y=uv;y'=u'v+v'u\\u'vtgx+v'utgx-uv=1\\u'vtgx+u(v'tgx-v)=1\\\begin{cases}v'tgx-v=0\\u'vtgx=1\end{cases}\\\frac{dv}{dx}tgx-v=0|*\frac{dx}{vtgx}\\\frac{dv}{v}-\frac{dx}{tgx}=0\\\frac{dv}{v}=\frac{dx}{tgx}\\\int\frac{dv}{v}=\int\frac{dx}{tgx}\\ln|v|=ln|sinx|\\v=sinx\\\frac{du}{dx}\frac{sin^2x}{cosx}=1|*\frac{cosxdx}{sin^2x}\\du=\frac{cosxdx}{sin^2x}\\\int du=\int \frac{cosxdx}{sin^2x}\\u=-\frac{1}{sinx}+C\\y=Csinx-1\\y(\frac{\pi}{6})=-\frac{1}{2}:\ -\frac{1}{2}=\frac{1}{2}C-1\\C=1$
$y=sinx-1$

$\frac{dy}{\sqrt x}+dx=\frac{dx}{\sqrt y}\\\frac{dy}{\sqrt x}=(\frac{1}{\sqrt y}-1)dx|*\frac{\sqrt x}{\frac{1}{\sqrt y}-1}\\\frac{\sqrt ydy}{1-\sqrt y}=\sqrt x dx\\\int\frac{\sqrt ydy}{1-\sqrt y}=\int\sqrt x dx\\\int\frac{\sqrt ydy}{1-\sqrt y}=2\int\frac{t^2dt}{1-t}=2\int(-t-1+\frac{1}{1-t})dt=-2(\frac{t^2}{2}+t+ln|1-t|)=\\=-y-\frac{\sqrt y}{2}-2ln|1-\sqrt y|\\y=t^2;t=\sqrt y;dy=2tdt\\-y-\frac{\sqrt y}{2}-2ln|1-\sqrt y|=\frac{2x^\frac{3}{2}}{3}+C$
$\frac{2x^\frac{3}{2}}{3}+y+\frac{\sqrt y}{2}+2ln|1-\sqrt y|=C$
А вот тут частное решение с данными параметрами не найти.