Question

Решить дифференциальное уравнение второго порядка.

xy ''- y' + 1/x = 0
​

Answer 1

$xy''-y'+\dfrac{1}{x}=0\ \ \Big|:x\ne 0\\\\y''-\dfrac{1}{x}\, y'=-\dfrac{1}{x^2}\ \ ,\ \ \ \ y'=p(x)\ \ ,\\\\p'-\dfrac{1}{x}\, p=-\dfrac{1}{x^2}\ \ ,\ \ \ p=uv\ ,\ \ p'=u'v =uv'\ \ ,\\\\u'v+uv'-\dfrac{1}{x}\, uv=-\dfrac{1}{x^2}\ \ \ \to \ \ \ \ u'v+u\cdot \Big(v'-\dfrac{1}{x}\, v\Big)=-\dfrac{1}{x^2}\\\\a)\ \ v'-\dfrac{1}{x}\, v=0\ \ ,\ \ \ \dfrac{dv}{dx}=\dfrac{v}{x}\ \ ,\ \ \ \int \dfrac{dv}{v}=\int \dfrac{dx}{x}\ \ ,\ \ ln|v|=ln|x|\\\\{}\qquad \boxed{\ v=x\ }$

$b)\ \ u'v=-\dfrac{1}{x^2}\ \ ,\ \ \dfrac{du}{dx}\cdot x=-\dfrac{1}{x^2}\ \ ,\ \ \ \dfrac{du}{dx}=-\dfrac{1}{x^3}\ \ ,\ \ \int du=-\int \dfrac{dx}{x^3}\ \ ,\\\\u=-\dfrac{x^{-2}}{-2}+C_1\ \ ,\ \ \ \boxed{\ u=\dfrac{1}{2x^2}+C_1\ }$

$c)\ \ p(x)=uv\ \ ,\ \ \ p(x)=x\cdot \Big(\dfrac{1}{2x^2}+C_1\Big)\ \ \ \to \ \ \ y'=x\cdot \Big(\dfrac{1}{2x^2}+C_1\Big)\ \ ,\\\\\dfrac{dy}{dx} =x\cdot \Big(\dfrac{1}{2x^2}+C_1\Big)\ \ ,\ \ \int dy=\int \Big(\dfrac{1}{2x}+C_1x\Big)\, dx\ \ ,\\\\\boxed{\ y=\dfrac{1}{2}\, ln|x|+C_1\cdot \dfrac{x^2}{2}+C_2\ }\ \ -\ otvet$