y^'=((4t^3+1))/((2y-6))

解答

y^{^{'}} = \frac{( 4 t ^{3} + 1 )}{( 2 y - 6 )}

y = 3 + t^{4} + t + c_{1} + 9, y = 3 - t^{4} + t + c_{1} + 9

求解步骤

y^{'} (t) = \frac{( 4 t ^{3} + 1 )}{( 2 y - 6 )}

求解可分离方程:

y = 3 + t^{4} + t + c_{1} + 9, y = 3 - t^{4} + t + c_{1} + 9

y = 3 + t^{4} + t + c_{1} + 9, y = 3 - t^{4} + t + c_{1} + 9

x \to - 1 lim (\frac{2 x}{∣ x ∣})

\int \frac{e ^{x^{\frac{1}{2}}}}{x ^{3}} d x

\int (- 2 T + x) \cdot e^{- i \cdot z \cdot x} d x

l a pl a ce t r an s f or m (t^{2}) \cdot (sin (2 \cdot t))

x \to 25 lim (\frac{\frac{1}{x} - \frac{1}{5}}{x - 25})