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

解答

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

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

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}) + c_{1})

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

t an g e n t y = x^{4} - 4 x^{3} + 2

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