x^'+5x^2=25

解答

x^{^{'}} + 5 x^{2} = 25

x = - \frac{5 e ^{105 t + 25 c_{1}}}{1 - e ^{25 (5 t + c_{1})}} - \frac{5}{1 - e ^{25 (5 t + c_{1})}}

求解步骤

x^{'} (t) + 5 x^{2} = 25

求解可分离方程:

x = - \frac{5 e ^{105 t + 25 c_{1}}}{1 - e ^{25 (5 t + c_{1})}} - \frac{5}{1 - e ^{25 (5 t + c_{1})}}

x = - \frac{5 e ^{105 t + 25 c_{1}}}{1 - e ^{25 (5 t + c_{1})}} - \frac{5}{1 - e ^{25 (5 t + c_{1})}}

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