Question

Сможете решить все кроме 3
Пожалуйста , помогите

Answer 1

$1)\; \; (x^2-1)y'+2xy^2=0\; \; ;\;\; y(0)=1\\\\y'=-\frac{2xy^2}{x^2-1}\; \; ,\; \; \int \frac{dy}{y^2}=-\int \frac{2x\, dx}{x^2-1}\; \; ,\; \; \frac{y^{-1}}{-1}=-ln|x^2-1|+C\; \; ,\\\\\frac{1}{y}=ln|x^2-1|+C\\\\y(0)=1:\; \; 1=ln|-1|+C\; ,\; \; 1=\underbrace {ln1}_{0}+C\; ,\; \; C=1\\\\\frac{1}{y}=ln|x^2-1|+1\; \; \Rightarrow \; \; y=\frac{1}{1+ln|x^2-1|}$

$2)\; \; (x+2y)\, dx-x\, dy=0\\\\\frac{dy}{dx}=\frac{x+2y}{x}\; \; ,\; \; y'=1+2\cdot \frac{y}{x}\\\\u=\frac{y}{x}\; ,\; \; y=ux\; ,\; \; y'=u'x+u\\\\u'x+u=1+2u\; \; ,\; \; u'x=1+u\; \; ,\; \; \frac{du}{dx}=\frac{1+u}{x}\\\\\int \frac{du}{1+u}=\int \frac{dx}{x}\; \; \Rightarrow \; \; ln|1+u|=ln|x|+lnC\\\\1+\frac{y}{x}=Cx\; ,\; \; \frac{y}{x}=Cx-1\; \; \Rightarrow \; \; \underline {y=x\cdot (Cx-1)}$

$4)\; \; 2y''-5y'+2y=0\; \; ;\; \; y(0)=1\; ,\; y'(0)=2\\\\2k^2-5k+2=0\; ,\; \; D=25-16=9\; ,\; \; k_1=2\; ,\; k_2=\frac{1}{2}\\\\\underline {y_{obshee}=C_1\cdot e^{2x}+C_2\cdot e^{\frac{x}{2}}}\\\\y(0)=1:\; \; 1=C_1+C_2\\\\y'(x)=C_1\cdot 2e^{2x}+C_2\cdot \frac{1}{2}\cdot e^{\frac{x}{2}}\\\\y'(0)=2:\; \; 2=2C_1+\frac{1}{2}C_2\; ,\; \; 4C_1+C_2=4\; ,\\\\\left \{ {{C_1+C_2=1} \atop {4C_1+C_2=4}} \right. \; \ominus \; \left \{ {{C_1+C_2=1} \atop {3C_1=3}} \right. \; \; \left \{ {{C_2=0} \atop {C_1=1}} \right.\\\\\underline {y_{chastnoe}=e^{2x}}$

$5)\; \; 2y'-\frac{x}{y}=\frac{xy}{x^2-1}\; \; \Big |\cdot y\\\\2yy'-x=\frac{xy^2}{x^2-1}\\\\Zamena:\; \; z=y^2\; \; \to \; \; z'=2yy'\; \; \Rightarrow \; \; \; z'-x=\frac{xz}{x^2-1}\; ,\\\\z'-\frac{x}{x^2-1}\cdot z=x\\\\z=uv\; ,\; \; z'=u'v+uv'\\\\u'v+uv'-\frac{x}{x^2-1}\cdot uv=x\\\\u'v+u\cdot (v'-\frac{x}{x^2-1})=x\; \; \Rightarrow \; \; \left \{ {{v'-\frac{x}{x^2-1}=0} \atop {xu'v=x}} \right.\\\\a)\; \; \frac{dv}{dx}=\frac{x}{x^2-1}\; \; ,\; \; \int \frac{dv}{v}=\frac{1}{2}\int \frac{2x\, dx}{x^2-1}\; \; ,\qquad \; \Big [\Big \; \int \frac{dt}{t}=ln|t|+C\; ]\\\\ln|v|=\frac{1}{2}\cdot ln|x^2-1|$

$v=\sqrt{x^2-1}\\\\b)\; \; u'\cdot \sqrt{x^2-1}=x\\\\\frac{du}{dx}=\frac{x}{\sqrt{x^2-1}}\; \; ,\; \; \int du=\frac{1}{2}\int \frac{2x\, dx}{\sqrt{x^2-1}}\; \; \; \; \; \; \Big [\; \int \frac{dt}{\sqrt{t}}=2\sqrt{t}+C\; \Big ]\\\\u=\frac{1}{2}\cdot 2\sqrt{x^2-1}+C=\sqrt{x^2-1}+C\\\\c)\; \; z=uv=\sqrt{x^2-1}\cdot (\sqrt{x^2-1}+C)\\\\\underline {y^2=x^2-1+C\cdot \sqrt{x^2-1}}$