(5y-2x)dy-2ydx=0

解答

(5 y - 2 x) d y - 2 y d x = 0

y = \frac{4 x + 16 x ^{2} + c _{1}}{10}, y = \frac{4 x - 16 x ^{2} + c _{1}}{10}

求解步骤

(5 y - 2 x) d y - 2 y d x = 0

求解恰当方程:

y = \frac{4 x + 16 x ^{2} + c _{1}}{10}, y = \frac{4 x - 16 x ^{2} + c _{1}}{10}

y = \frac{4 x + 16 x ^{2} + c _{1}}{10}, y = \frac{4 x - 16 x ^{2} + c _{1}}{10}

\frac{\partial}{\partial x} (\frac{3 x - 5 y}{3 x + 5 y})

\frac{\partial}{\partial y} (e^{x y} + 2 x - 5 y)

t an g e n t y = 2 x^{3} - 5 x + 1, (1, - 2)

\int x 1 - x^{2} d x

t an g e n t f (x) = \frac{10 x}{1 + x ^{2}}, (3, 3)