(dy)/(dt)+2(t+1)y^2=0,y(0)=-1/8

解答

\frac{d y}{d t} + 2 (t + 1) y^{2} = 0, y (0) = - \frac{1}{8}

y = - \frac{1}{- t ^{2} - 2 t + 8}

求解步骤

\frac{d y}{d t} + 2 (t + 1) y^{2} = 0

求解可分离方程:

y = - \frac{1}{- t ^{2} - 2 t + 8}

y = - \frac{1}{- t ^{2} - 2 t + 8}

y^{^{'}} = \frac{ln ( x )}{x y}, y (1) = 2

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

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

\frac{d h}{d t} = 1 - 0.5 h

\frac{( x ^{2} - 9 )}{x} \frac{d y}{d x} + y = 0, y (5) = 1