证明函数u(x,y)=f(y/x^2)*x^n满足x*(∂u/∂x)+2y*(∂u/
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证明函数u(x,y)=f(y/x^2)*x^n满足x*(∂u/∂x)+2y*(∂u/∂y)=nu
![证明函数u(x,y)=f(y/x^2)*x^n满足x*(∂u/∂x)+2y*(∂u/](/uploads/image/z/18923396-68-6.jpg?t=%E8%AF%81%E6%98%8E%E5%87%BD%E6%95%B0u%28x%2Cy%29%3Df%28y%2Fx%5E2%29%2Ax%5En%E6%BB%A1%E8%B6%B3x%2A%EF%BC%88%26%238706%3Bu%2F%26%238706%3Bx%29%2B2y%2A%28%26%238706%3Bu%2F)
(∂u/∂x)=(n)f(y/x^2)x^(n-1)+(-2y)(x^n)(x^-3)f'(y/x^2)
(∂u/∂y)=(x^n)(x^-2)f'(y/x^2)
x(∂u/∂x)=n(x^n)f(y/x^2)-2y(x^(n-2))f'(y/x^2)
2y(∂u/∂y)=2y(x^(n-2))f'(y/x^2)
所以x*(∂u/∂x)+2y*(∂u/∂y)=nf(y/x^2)*x^n=nu
(∂u/∂y)=(x^n)(x^-2)f'(y/x^2)
x(∂u/∂x)=n(x^n)f(y/x^2)-2y(x^(n-2))f'(y/x^2)
2y(∂u/∂y)=2y(x^(n-2))f'(y/x^2)
所以x*(∂u/∂x)+2y*(∂u/∂y)=nf(y/x^2)*x^n=nu
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