Question

Решите пожалуйста методом Бернули
ху'- 4у =- х +1

Answer 1

$xy'-4y=-x+1\\\\y'-\frac{4y}{x}=\frac{-x+1}{x}\\\\y=uv\; ,\; \; y'=u'v+uv'\\\\u'v+uv'-\frac{4uv}{x}=\frac{1-x}{x}\\\\u'v+u\cdot (v'-\frac{4v}{x})=\frac{1-x}{x}\\\\1)\; \; \frac{dv}{dx}=\frac{4v}{x}\; \; ,\; \; \int \frac{dv}{v}=\int \frac{4\, dx}{x}\; \; ,\; \; ln|v|=4ln|x|\; ,\; \; v=x^4\\\\2)\; \; u'\cdot x^4=\frac{1-x}{x}\; \; ,\; \; \frac{du}{dx}=\frac{1-x}{x^5}\; \; ,\; \; \int du=\int (\frac{1}{x^5}-\frac{1}{x^4})\, dx\; ,\\\\u=\frac{x^{-4}}{-4}-\frac{x^{-3}}{-3}+C\\\\u=-\frac{1}{4x^4}+\frac{1}{3x^3}+C$

$3)\; \; y=x^4\cdot (-\frac{1}{4x^4}+\frac{1}{3x^3}+C)=Cx^4+\frac{x}{3}-\frac{1}{4}$