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已知(x+ay)dx+ydy / (x+y)^2为某函数的全微分,则a=?
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若f(x,y)dx+g(x,y)dy为全微分
需满足
∂f(x,y)/∂y=∂g(x,y)/∂x
∂[(x+ay)/(x+y)²]/∂y
=(ax-2x-ay)/(x+y)³
∂[y/(x+y)²]/∂x
=-2y/(x+y)³
故(ax-2x-ay)/(x+y)³=-2y/(x+y)³
(a-2)(x-y)/(x+y)³=0
即a=2
需满足
∂f(x,y)/∂y=∂g(x,y)/∂x
∂[(x+ay)/(x+y)²]/∂y
=(ax-2x-ay)/(x+y)³
∂[y/(x+y)²]/∂x
=-2y/(x+y)³
故(ax-2x-ay)/(x+y)³=-2y/(x+y)³
(a-2)(x-y)/(x+y)³=0
即a=2
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若f(x,y)dx+g(x,y)dy为全微分
需满足
?f(x,y)/?y=?g(x,y)/?x
?[(x+ay)/(x+y)sup2;]/?y
=(ax-2x-ay)/(x+y)sup3;
?[y/(x+y)sup2;]/?x
=-2y/(x+y)sup3;
故(ax-2x-ay)/(x+y)sup3;=-2y/(x+y)sup3;
(a-2)(x-y)/(x+y)sup3;=0
即a=2
需满足
?f(x,y)/?y=?g(x,y)/?x
?[(x+ay)/(x+y)sup2;]/?y
=(ax-2x-ay)/(x+y)sup3;
?[y/(x+y)sup2;]/?x
=-2y/(x+y)sup3;
故(ax-2x-ay)/(x+y)sup3;=-2y/(x+y)sup3;
(a-2)(x-y)/(x+y)sup3;=0
即a=2
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