y^'=(e^{x-y})/(1+e^x),y(1)=0

解答

y^{^{'}} = \frac{e ^{x - y}}{1 + e ^{x}}, y (1) = 0

y = ln (ln (1 + e^{x}) + 1 - ln (1 + e))

求解步骤

y^{'} (x) = \frac{e ^{x - y}}{1 + e ^{x}}

求解可分离方程:

y = ln (ln (1 + e^{x}) + 1 - ln (1 + e))

y = ln (ln (1 + e^{x}) + 1 - ln (1 + e))

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