(dy)/(dt)= t/(t^2y+y),y(0)=4

解答

\frac{d y}{d t} = \frac{t}{t ^{2} y + y}, y (0) = 4

y = ln (t^{2} + 1) + 16

求解步骤

\frac{d y}{d t} = \frac{t}{t ^{2} y + y}

求解可分离方程:

y = ln (t^{2} + 1) + 16

y = ln (t^{2} + 1) + 16

(x^{2} + 1) (tan (y)) y^{^{'}} = x

y^{^{'}} = (1 + 4 x) y, y (0) = 1

(x^{- 2} y^{- 5}) d x + (x^{- 3} y^{- 4}) d y = 0

\frac{d V}{d t} = 20 - \frac{8}{100} V

\frac{d y}{d t} = \frac{2 t}{1 + 2 y}