(2xy+3y^2)dx=(2xy+x^2)dy

解答

(2 x y + 3 y^{2}) d x = (2 x y + x^{2}) d y

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

求解步骤

(2 x y + 3 y^{2}) d x = (2 x y + x^{2}) d y

使用替代法求解:

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

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

x \to 2 lim (\frac{x ^{3} - 8}{2 - x})

\int \frac{u ^{2}}{u ^{2} + 1} d u

k = 1 \sum \infty \frac{k}{2 ^{k}}

\frac{d}{d x} (cosh (x^{2}))

d er i v a t i v e 5 (9 x^{7} + 3 x^{4} - 2)^{3}