y^'= 5/(y+2)

解答

y^{^{'}} = \frac{5}{y + 2}

y = \frac{- 4 + 16 - 4 ( - c _{1} - 10 t )}{2}, y = \frac{- 4 - 16 - 4 ( - c _{1} - 10 t )}{2}

求解步骤

y^{'} (t) = \frac{5}{y + 2}

求解可分离方程:

y = \frac{- 4 + 16 - 4 ( - c _{1} - 10 t )}{2}, y = \frac{- 4 - 16 - 4 ( - c _{1} - 10 t )}{2}

y = \frac{- 4 + 16 - 4 ( - c _{1} - 10 t )}{2}, y = \frac{- 4 - 16 - 4 ( - c _{1} - 10 t )}{2}

\frac{d}{d x} ((5 x^{6} + 8 x^{3})^{4})

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t an g e n t y = x + x, (49, 56)

\int_{1}^{25} \frac{2}{x} d x