Question

Решить уравнение:
x(y"+1)=-y'

Answer 1

$displaystyle x(y''+1)=-y'\z=y';z'=y''\x(z'+1)=-z\xz'+z=-x\xz'+z=0\frac{dz}{z}=-frac{dx}{x}\ln|z|=-ln|x|+ln|C|\z=frac{C(x)}{x};z'=frac{C'(x)}{x}-frac{C(x)}{x^2}\C'(x)-frac{C(x)}{x}+frac{C(x)}{x}=-x\C'(x)=-xto C(x)=-frac{x^2}{2}+C_1\z=-frac{x}{2}+frac{C_1}{x}=y'\y=int(-frac{x}{2}+frac{C_1}{x})dx=-frac{x^2}{4}+C_1ln|x|+C_2$