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](/uploads/image/z/14036220-36-0.jpg?t=%E7%94%A8a%EF%BC%8Fb%EF%BC%9DIn+a%EF%BC%8DIn+b%E8%AF%81%E6%98%8E%E8%AE%A1%E7%AE%97%E7%94%A8a%EF%BC%8Fb%EF%BC%9DIn+a%EF%BC%8DIn+b%2C%E8%AF%81%E6%98%8E%2C%E5%A6%82%E6%9E%9Cy%EF%BC%9DIn%EF%BC%881%EF%BC%8Bx%EF%BC%8F1%EF%BC%8Dx%EF%BC%89%E8%BF%99%E4%B8%AA%E5%BC%8F%E5%AD%90%EF%BC%8E%EF%BC%882x%EF%BC%8F1%EF%BC%8Dx%5E2%EF%BC%89dy%2Fdx%EF%BC%8Dd%5E2y%EF%BC%8Fdx%5E2%EF%BC%9D0)
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
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
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
我没时间算了