derivative of 3y^2+2x^3-14=0

解答

\frac{d}{d x} (3 y^{2}) + 2 x^{3} - 14 = 0

y = \frac{- x ^{4} + 2 ( 14 x + c _{1} )}{6}, y = - \frac{- x ^{4} + 2 ( 14 x + c _{1} )}{6}

求解步骤

\frac{d}{d x} (3 y^{2}) + 2 x^{3} - 14 = 0

整理

(3 y^{2})^{'} : (3 y^{2})^{'} = - 2 x^{3} + 14

若

f^{^{'}} (x) = g (x)

则

f (x) = \int g (x) d x

3 y^{2} = \int - 2 x^{3} + 14 d x

\int - 2 x^{3} + 14 d x = - \frac{x ^{4}}{2} + 14 x + c_{1}

3 y^{2} = - \frac{x ^{4}}{2} + 14 x + c_{1}

整理

y : y = \frac{- x ^{4} + 2 ( 14 x + c _{1} )}{6}, y = - \frac{- x ^{4} + 2 ( 14 x + c _{1} )}{6}

y = \frac{- x ^{4} + 2 ( 14 x + c _{1} )}{6}, y = - \frac{- x ^{4} + 2 ( 14 x + c _{1} )}{6}

x^{^{'}} = x^{3}, x (0) = 1

1 + t = 2 t x \frac{d x}{d t}

2 y (x^{2} + 1) d y = x d x

y^{2} e^{x} + 3 y e^{2 x} \frac{d y}{d x} = 2 e^{x}, y (0) = 1

\frac{x ^{2}}{y ^{2} - 1} y^{^{'}} = \frac{3 x ^{3}}{y}