2ty^'-3y=0

解答

2 t y^{^{'}} - 3 y = 0

y = c_{1} t^{\frac{3}{2}}

求解步骤

2 t y^{'} (t) - 3 y = 0

求解可分离方程:

y = c_{1} t^{\frac{3}{2}}

y = c_{1} t^{\frac{3}{2}}

f (x) = \frac{7}{x ^{4}}, at x = - 3

y = x + 2, (- 1, 1)

\frac{d y}{d x} = \frac{y ^{2} + x y + x ^{2}}{x ^{2}}, y (1) = - 1

x \to - 4 + lim (\frac{- 2}{x ^{2} - 16})

\frac{\partial}{\partial x} (ln (\frac{x}{4}))