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

解答

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

y = 1 + t^{3} + 2 t^{2} + 2 t + c_{1} + 1, y = 1 - t^{3} + 2 t^{2} + 2 t + c_{1} + 1

求解步骤

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

求解可分离方程:

y = 1 + t^{3} + 2 t^{2} + 2 t + c_{1} + 1, y = 1 - t^{3} + 2 t^{2} + 2 t + c_{1} + 1

y = 1 + t^{3} + 2 t^{2} + 2 t + c_{1} + 1, y = 1 - t^{3} + 2 t^{2} + 2 t + c_{1} + 1

y tan (x) \frac{d y}{d x} = (4 + y^{2}) sec (x^{2})

\frac{d y}{d x} = c \cdot y

tan (x) d y + 2 y d x = 0

y^{^{'}} = 7 x y

\frac{1}{y} d y = d x