Question

Найти y’ и y’’

y2 = x + ln(y/x)

Answer 1

$y^2=x+ln\frac{y}{x}\\\\2yy'=1+\frac{x}{y}\cdot \frac{1}{x}\cdot y'\\\\y'(2y-\frac{1}{y})=1\\\\y'\cdot \frac{2y^2-1}{y} =1\\\\y'=\frac{y}{2y^2-1}$