Question

Помогите решить!!!! Очень надо

Answer 1

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

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

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

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

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

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

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

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