Tôi nghĩ đề cả bạn phải là
Cho $u= f(\frac{y}{x}) + x.h(\frac{y}{x})$ ,trong đó f và h là những hàm có đạo hàm cấp hai.
Tính: $A = \frac{y}{x}*\frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial x\partial y}$
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Ta có
$\frac{\partial u}{\partial y}=\frac{1}{x}.f^{'}\left ( \frac{y}{x} \right )+h^{'}\left ( \frac{y}{x} \right )$
$\frac{\partial ^{2}u}{\partial y^{2}}=\frac{1}{x^{2}}.f^{''}\left ( \frac{y}{x} \right )+\frac{1}{x}.h^{''}\left ( \frac{y}{x} \right )$
$\frac{\partial ^{2}u}{\partial x\partial y}=-\frac{1}{x^{2}}.f^{'}\left ( \frac{y}{x} \right )+\frac{1}{x}.\frac{-y}{x^{2}}.f^{''}\left ( \frac{y}{x} \right )-\frac{y}{x^{2}}h^{''}\left ( \frac{y}{x} \right )$
Suy ra
$A=\frac{y}{x}.\frac{\partial ^{2}u}{\partial y^{2}}+\frac{\partial ^{2}u}{\partial x\partial y}$
$=\frac{y}{x^{3}}.f^{''}\left ( \frac{y}{x} \right )+\frac{y}{x^{2}}.h^{''}\left ( \frac{y}{x} \right )-\frac{1}{x^{2}}f^{'}\left ( \frac{y}{x} \right )-\frac{y}{x^{3}}.f^{''}\left ( \frac{y}{x} \right )-\frac{y}{x^{2}}.h^{''}\left ( \frac{y}{x} \right )$
$=-\frac{1}{x^{2}}.f^{'}\left ( \frac{y}{x} \right )$
Bài viết đã được chỉnh sửa nội dung bởi vo van duc: 21-01-2013 - 08:00