y^'= t/(1+t^2)*e^y

解答

y^{^{'}} = \frac{t}{1 + t ^{2}} \cdot e^{y}

y = - ln (- \frac{1}{2} ln (1 + t^{2}) - c_{1})

求解步骤

y^{'} (t) = \frac{t}{1 + t ^{2}} \cdot e^{y}

求解可分离方程:

y = - ln (- \frac{1}{2} ln (1 + t^{2}) - c_{1})

y = - ln (- \frac{1}{2} ln (1 + t^{2}) - c_{1})

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