Question

Найти y' и y''
найти y' и y''​

Answer 1

$y^2=x+ln\dfrac{y}{x}\\\\2yy'=1+\dfrac{x}{y}\cdot \dfrac{y'x-y}{x^2}\ \ \ ,\ \ \ \2yy'=1+\dfrac{y'x-y}{xy}\ \ \ ,\ \ \ 2yy'=1+\dfrac{y'}{y}-\dfrac{1}{x}\\\\\\2yy'-\dfrac{y'}{y}=1-\dfrac{1}{x}\ \ \ ,\ \ \ y'\cdot \Big(2y-\dfrac{1}{y}\Big)=\dfrac{x-1}{x}\ \ ,\ \ \ y'\cdot \dfrac{2y^2-1}{y}=\dfrac{x-1}{x}\ ,\\\\\\\boxed{\ y'=\dfrac{y\cdot (x-1)}{x\cdot (2y^2-1)}\ }$

$y''=\dfrac{\Big(y'(x-1)+y\Big)\cdot x\cdot (2y^2-1)-y(x-1)\cdot \Big(2y^2-1+x\cdot 4yy'\Big)}{x^2(2y^2-1)^2}=\\\\\\=\dfrac{\Big(\dfrac{y\cdot (x-1)^2}{x\cdot (2y^2-1)}+y\Big)\cdot x\cdot (2y^2-1)-y\cdot (x-1)\cdot \Big(2y^2-1+4xy\cdot \dfrac{y\cdot (x-1)}{x\cdot (2y^2-1)}\Big)}{x^2\cdot (2y^2-1)^2}=\\\\\\=\dfrac{\Big(y(x-1)^2+xy(2y^2-1)\Big)\cdot x(2y^2-1)-y(x-1)\cdot \Big(x(2y^2-1)^2+4y^2(x-1)\Big)}{x^3(2y^2-1)^3}$