Question

Помогите решить!!!! Очень надо

Answer 1

$frac{partial u}{partial x}=(x+y)`_{x}cdot e^{frac{x}{y}}+(x+y)cdot ( e^{frac{x}{y}})`_{x}=e^{frac{x}{y}}+(x+y)cdot ( e^{frac{x}{y}})cdot (frac{x}{y})`_{x}=\ \ =e^{frac{x}{y}}+(x+y)cdot e^{frac{x}{y}}cdot(frac{1}{y})=e^{frac{x}{y}}cdot (2+frac{x}{y})$

$frac{partial u}{partial y}=(x+y)`_{y}cdot e^{frac{x}{y}}+(x+y)cdot ( e^{frac{x}{y}})`_{y}=e^{frac{x}{y}}+(x+y)cdot ( e^{frac{x}{y}})cdot (frac{x}{y})`_{y}=\ \ =e^{frac{x}{y}}+(x+y)cdot e^{frac{x}{y}}cdot(-frac{x}{y^2})=e^{frac{x}{y}}cdot (1-frac{x^2}{y^2}-frac{x}{y})$

$du=frac{x}{y} dx+frac{x}{y} dy=e^{frac{x}{y}}cdot(2+frac{x}{y})dx+e^{frac{x}{y}}cdot(1-frac{x^2}{y^2}-frac{x}{y})dy$

$frac{partial ^2u}{partial x^2}=(e^{frac{x}{y}}cdot(2+frac{x}{y})_{x}= e^{frac{x}{y}}cdot (frac{1}{y})cdot(2+frac{x}{y})+ e^{frac{x}{y}}cdot (frac{1}{y})= e^{frac{x}{y}}cdot (frac{3}{y}+frac{x}{y^2})$

$frac{partial^2 u}{partial x partial y}=(e^{frac{x}{y}}cdot(2+frac{x}{y}))`_{y}=e^{frac{x}{y}}cdot (-frac{x}{y^2})cdot (2+frac{x}{y})+e^{frac{x}{y}}cdot(-frac{x}{y^2})=e^{frac{x}{y}}cdot (-frac{3x}{y^2}-frac{x^2}{y^3}$

$frac{partial ^2u}{partial y^2}=(e^{frac{x}{y}}cdot(1-frac{x}{y^2}-frac{x}{y}))_{y}=e^{frac{x}{y}}cdot (-frac{x}{y^2})cdot (1-frac{x}{y^2}-frac{x}{y})+ e^{frac{x}{y}}cdot (frac{2x}{y^3}+frac{x}{y^2})$

$d^2u=frac{partial^2 u}{partial x^2}dx^2+2cdot frac{partial^2 u}{partial xpartial y}dxdy+frac{partial^2 u}{partial y^2}dy^2$

$d^2u=e^{frac{x}{y}}((2+frac{x}{y})dx^2+2cdot cdot (frac{x^2}{y^4}+frac{x^2}{y^3}+frac{2x}{y^3}) dxdy+(frac{2x}{y^3}+frac{x}{y^2})cdot dy^2)$