用a/b=In a-In b证明计算用a/b=In a-In b,证明,如果y=In(1+x/1-x)这个式子.(2x/1-x^2)dy/dx-d^2y/dx^2=0
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用a/b=In a-In b证明计算
用a/b=In a-In b,证明,如果y=In(1+x/1-x)
这个式子.
(2x/1-x^2)dy/dx-d^2y/dx^2=0
Since a/b = ln a - ln b
So y = ln ((1+x)/(1-x)) = ln (1+x) - ln (1-x)
dy/dx = d[ln(1+x)]/dx - d[ln(1-x)]/dx
= 1/(1+x) * d(1+x)/dx = 1/(1-x) * d(1-x)/dx
= 1/(1+x) - 1/(1-x) * (-1)
= 1/(1+x) + 1/(1-x) = (1-x+1+x)/((1+x)(1-x)) = 2/(1-x^2)
Let 1-x^2 = u
d^2y/dx^2 = d[2/(1-x^2)]/dx =d(2/u)/dx = (-1)*2*u^(-2) * du/dx
=(-2) * [1/(1-x^2)^2] * (-2x) = 4x/(1-x^2)^2
Therefore,
[2x/(1-x^2)] dy/dx = [2x(1-x^2)] * [2/(1-x^2)] = 4x/(1-x^2)^2 = d^2y/dx^2
Yeah!
令y'=dy/dx=p,
则y''=d^2y/dx^2=dp/dx
∴(2x/1-x^2)dy/dx-d^2y/dx^2
=p·(2x/1-x^2)-dp/dx=0;
分离变量:dp/p=dx/(2x/1-x^2)
然后接着解微分方程就行了
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用a/b=In a-In b证明计算用a/b=In a-In b,证明,如果y=In(1+x/1-x)这个式子.(2x/1-x^2)dy/dx-d^2y/dx^2=0
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