(2x-1)dx+(7y+4)dy=0

解答

(2 x - 1) d x + (7 y + 4) d y = 0

y = \frac{- 8 + 64 - 28 ( - c _{1} - 2 x + 2 x ^{2} )}{14}, y = \frac{- 8 - 64 - 28 ( - c _{1} - 2 x + 2 x ^{2} )}{14}

求解步骤

(2 x - 1) d x + (7 y + 4) d y = 0

求解可分离方程:

y = \frac{- 8 + 64 - 28 ( - c _{1} - 2 x + 2 x ^{2} )}{14}, y = \frac{- 8 - 64 - 28 ( - c _{1} - 2 x + 2 x ^{2} )}{14}

y = \frac{- 8 + 64 - 28 ( - c _{1} - 2 x + 2 x ^{2} )}{14}, y = \frac{- 8 - 64 - 28 ( - c _{1} - 2 x + 2 x ^{2} )}{14}

\int \frac{3 x ^{2} + 4 x + 5}{x} d x

\frac{\partial}{\partial y} (16 - 4 x^{2} - y^{2})

d er i v a t i v e g (x) = x^{2} (1 - 5 x)

\frac{\partial}{\partial x} (2 z^{3} - 3 (x^{2} + y^{2}) z)

x \to - 8 + lim (\frac{x}{x + 8})