sinx? |
cosX? |
f'(y)-f'(y)=-2f(0)siny f'(y)+f'(y)=2f'(0)cosy f'(y)=-f(0)siny+f'(0)cosy f(y)=f(0)cosy+f'(0)siny f(0)=2f(0) --> f(0)=0 f(y) =f'(0)siny --> f(x)=Asinx, A--any real number |
f '(y)-f '(-y)=-2f(0)siny f '(y)+f '(y)=2f '(0)cosy f '(y)=-f(0)siny+f '(0)cosy f(y)=f(0)cosy+f'(0)siny So f(x) = Asin(x+x0) where A and x0 are any real numbers |
好象求导数的时候有点不太准确 |
已知实函数f(x)满足:f(x+y)+f(x-y)=2f(x)*cosy,求f(x)
The const should be 0 to satisfy f(x+y)+f(x-y)=2f(x)*cosy. |
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