Решение.
[tex]\bf u=e^{^{\frac{x}{y}}}\ \ ,\ \ \ \ \dfrac{\partial }{\partial x}\Big(x\cdot \dfrac{\partial u}{\partial x}\Big)-y^2\cdot \dfrac{\partial ^2u}{\partial y^2}=0[/tex]
Вычислим частные производные .
[tex]\bf \dfrac{\partial u}{\partial x}=e^{^{\frac{x}{y} }}\cdot \dfrac{1}{y}\\\\\\\dfrac{\partial }{\partial x}\Big(x\cdot \dfrac{\partial u}{\partial x}\Big)=\dfrac{\partial }{\partial x}\Big(e^{^{\frac{x}{y}}}\cdot \dfrac{x}{y}\Big)=e^{^{\frac{x}{y}}}\cdot \dfrac{1}{y}\cdot \dfrac{x}{y}+e^{^{\frac{x}{y}}}\cdot \dfrac{1}{y}=e^{^{\frac{x}{y}}}\cdot \dfrac{1}{y}\cdot \Big(\dfrac{x}{y}+1\Big)\\\\\\\dfrac{\partial u}{\partial y}=e^{^{\frac{x}{y}}}\cdot \dfrac{-x}{y^2}=-e^{^{\frac{x}{y}}}\cdot \dfrac{x}{y^2}[/tex]
[tex]\bf \dfrac{\partial ^2u}{\partial y^2}=-e^{^{\frac{x}{y}}}\cdot \dfrac{-x}{y^2}\cdot \dfrac{x}{y^2}-e^{^{\frac{x}{y}}}\cdot \dfrac{-x\cdot 2y}{y^4}=e^{^{\frac{x}{y}}}\cdot \dfrac{x^2}{y^4}+e^{^{\frac{x}{y}}}\cdot \dfrac{2x}{y^3}=e^{^{\frac{x}{y}}}\cdot \dfrac{x}{y^3}\cdot \Big(\dfrac{x}{y}+2\Big)[/tex]
Подставим найденные выражения в функцию .
[tex]\bf \dfrac{\partial }{\partial x}\Big(x\cdot \dfrac{\partial u}{\partial x}\Big)-y^2\cdot \dfrac{\partial ^2u}{\partial y^2}=e^{^{\frac{x}{y}}}\cdot \dfrac{1}{y}\cdot \Big(\dfrac{x}{y}+1\Big)-y^2\cdot e^{^{\frac{x}{y}}}\cdot\dfrac{x}{y^3}\cdot \Big(\dfrac{x}{y}+2\Big)=[/tex]
[tex]\bf =\dfrac{e^{^{\frac{x}{y}}}}{y}\cdot \Big(\dfrac{x}{y}+1\Big)-\dfrac{e^{^{\frac{x}{y}}}\cdot x}{y}\cdot \Big(\dfrac{x}{y}+2\Big)=\dfrac{e^{^{\frac{x}{y}}}\cdot x}{y^2}+\dfrac{e^{^{\frac{x}{y}}}}{y}-\dfrac{e^{^{\frac{x}{y}}}\cdot x^2}{y^2}-\dfrac{2\, \, x\, e^{^{\frac{x}{y}}}}{y}=[/tex]
[tex]\bf =\dfrac{e^{^{\frac{x}{y}}}}{y}\cdot \Big(\dfrac{x}{y}+1-\dfrac{x^2}{y}-2x\Big)\ne 0[/tex]
Заданная функция не является решением уравнения .
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Решение.
[tex]\bf u=e^{^{\frac{x}{y}}}\ \ ,\ \ \ \ \dfrac{\partial }{\partial x}\Big(x\cdot \dfrac{\partial u}{\partial x}\Big)-y^2\cdot \dfrac{\partial ^2u}{\partial y^2}=0[/tex]
Вычислим частные производные .
[tex]\bf \dfrac{\partial u}{\partial x}=e^{^{\frac{x}{y} }}\cdot \dfrac{1}{y}\\\\\\\dfrac{\partial }{\partial x}\Big(x\cdot \dfrac{\partial u}{\partial x}\Big)=\dfrac{\partial }{\partial x}\Big(e^{^{\frac{x}{y}}}\cdot \dfrac{x}{y}\Big)=e^{^{\frac{x}{y}}}\cdot \dfrac{1}{y}\cdot \dfrac{x}{y}+e^{^{\frac{x}{y}}}\cdot \dfrac{1}{y}=e^{^{\frac{x}{y}}}\cdot \dfrac{1}{y}\cdot \Big(\dfrac{x}{y}+1\Big)\\\\\\\dfrac{\partial u}{\partial y}=e^{^{\frac{x}{y}}}\cdot \dfrac{-x}{y^2}=-e^{^{\frac{x}{y}}}\cdot \dfrac{x}{y^2}[/tex]
[tex]\bf \dfrac{\partial ^2u}{\partial y^2}=-e^{^{\frac{x}{y}}}\cdot \dfrac{-x}{y^2}\cdot \dfrac{x}{y^2}-e^{^{\frac{x}{y}}}\cdot \dfrac{-x\cdot 2y}{y^4}=e^{^{\frac{x}{y}}}\cdot \dfrac{x^2}{y^4}+e^{^{\frac{x}{y}}}\cdot \dfrac{2x}{y^3}=e^{^{\frac{x}{y}}}\cdot \dfrac{x}{y^3}\cdot \Big(\dfrac{x}{y}+2\Big)[/tex]
Подставим найденные выражения в функцию .
[tex]\bf \dfrac{\partial }{\partial x}\Big(x\cdot \dfrac{\partial u}{\partial x}\Big)-y^2\cdot \dfrac{\partial ^2u}{\partial y^2}=e^{^{\frac{x}{y}}}\cdot \dfrac{1}{y}\cdot \Big(\dfrac{x}{y}+1\Big)-y^2\cdot e^{^{\frac{x}{y}}}\cdot\dfrac{x}{y^3}\cdot \Big(\dfrac{x}{y}+2\Big)=[/tex]
[tex]\bf =\dfrac{e^{^{\frac{x}{y}}}}{y}\cdot \Big(\dfrac{x}{y}+1\Big)-\dfrac{e^{^{\frac{x}{y}}}\cdot x}{y}\cdot \Big(\dfrac{x}{y}+2\Big)=\dfrac{e^{^{\frac{x}{y}}}\cdot x}{y^2}+\dfrac{e^{^{\frac{x}{y}}}}{y}-\dfrac{e^{^{\frac{x}{y}}}\cdot x^2}{y^2}-\dfrac{2\, \, x\, e^{^{\frac{x}{y}}}}{y}=[/tex]
[tex]\bf =\dfrac{e^{^{\frac{x}{y}}}}{y}\cdot \Big(\dfrac{x}{y}+1-\dfrac{x^2}{y}-2x\Big)\ne 0[/tex]
Заданная функция не является решением уравнения .