Question

Найти общее решение дифференциального уравнения.

Answer 1

Ответ:

$\displaystyle xy'''+y''=x+1\\\\y'''+\frac{1}{x}\, y''=\frac{x+1}{x}\ \ ,\ \ \ \ \ \ y'=t(x)\ \ ,\ \ y''=t'(x)\ \ ,\ \ y'''=t''(x)\ \ \Rightarrow \\\\t''+\frac{1}{x}\, t'=\frac{x+1}{x}\ \ ,\qquad \ \ t'=p(x)\ \ ,\ \ t''=p'(x)\\\\p'+\frac{1}{x}\, p=\frac{x+1}{x}\\\\p=uv\ ,\ \ p'=u'v+uv'\\\\u'v+uv'+\frac{uv}{x}=\frac{x+1}{x}\ \ ,\ \ \ \ u'v+u\, (\, v'+\frac{v}{x}\ )=\frac{x+1}{x}\ \ ,\\\\\\a)\ \ v'+\frac{v}{x}=0\ \ ,\ \ \ \int \frac{dv}{v}=-\int \frac{dx}{x}\ \ ,\ \ \ lnv=-lnx\ \ ,\ \ \ v=\frac{1}{x}$

$\displaystyle b)\ \ u'v=\frac{x+1}{x}\ \ ,\ \ \ \frac{du}{dx}\cdot \frac{1}{x}=\frac{x+1}{x}\ \ ,\ \ \ \int du=\int (x+1)\, dx\ \ ,\\\\u=\frac{x^2}{2}+x+C_1\\\\\\c)\ \ p(x)=uv=\frac{1}{x}\cdot \Big(\frac{x^2}{2}+x+C_1\Big)=\frac{x}{2}+1+\frac{C_1}{x}\ \ \ ,\ \ \ \ p=t'(x)\ \ \Rightarrow \\\\\\d)\ \ \frac{dt}{dx}=\frac{x}{2}+\frac{C_1}{x}+1\ \ ,\ \ \int dt=\int \Big(\frac{x}{2}+\frac{C_1}{x}+1\Big)\, dx\ \ ,\\\\\\t=\frac{x^2}{4}+C_1\, ln|x|+x+C_2\ \ ,\ \ \ \ t=y'(x)\ \ \Rightarrow$

$\displaystyle e)\ \ \frac{dy}{dx}=\frac{x^2}{4}+C_1\, ln|x|+x+C_2\ \ \ , \ \ \ \int dy=\int \Big(\frac{x^2}{4}+C_1\, lnx+x+C_2\Big)\, dx\ \ ,\\\\\\\boxed{\ y=\frac{x^3}{12}+C_1\cdot (x\cdot lnx-x)+\frac{x^2}{2}+C_2x+C_3\ }$