Question

Решить интегральное уравнение xy'-y=y^3

Answer 1

$xy'-y=y^3\; |:x\\\\y'- \frac{y}{x}=\frac{y^3}{x}\\\\y=uv\; ,\; \; y'=u'v+uv'\\\\u'v+uv'- \frac{uv}{x}=\frac{(uv)^3}{x}\\\\u'v+u(\underbrace {v'- \frac{v}{x}}_{0})=\frac{(uv)^3}{x}\\\\a)\; \; \frac{dv}{dx}-\frac{v}{x}=0\; \; ,\; \; \frac{dv}{dx}=\frac{v}{x}\; \; ,\; \; \int \frac{dv}{v}=\int \frac{dx}{x}\\\\ln|v|=ln|x|\; \; \; \to \; \; \; v=x\\\\b)\; \; u'\cdot x= \frac{(ux)^3}{x}\; \\\\\frac{du}{dx}\cdot x=\frac{u^3x^3}{x} \; \; ,\; \; \int \frac{du}{u^3}=\int x\, dx\\\\\frac{u^{-2}}{-2}=\frac{x^2}{2}-\frac{C}{2}$

$- \frac{1}{2u^2}=\frac{x^2}{2}-\frac{C}{2}\\\\-\frac{1}{u^2}=x^2-C\\\\u^2=\frac{1}{C-x^2}\\\\u=\pm \frac{1}{\sqrt{C-x^2}}\\\\c)\; \; \; y=\pm \frac{x}{\sqrt{C-x^2}}$