y^'=y(1-y),y(0)=6

解答

y^{^{'}} = y (1 - y), y (0) = 6

y = - \frac{6 e ^{t}}{5 - 6 e ^{t}}

求解步骤

y^{'} (t) = y (1 - y)

求解可分离方程:

y = - \frac{6 e ^{t}}{5 - 6 e ^{t}}

y = - \frac{6 e ^{t}}{5 - 6 e ^{t}}

y^{^{'}} = 1 - y^{2} sin (x)

x \to 0 + lim (\frac{x e ^{\frac{1}{x}}}{2})

\int 9 x sin^{2} (x) d x

\int \frac{( 2 x - 3 )}{( x ^{2} + 2 x - 3 ) ^{\frac{3}{2}}} d x

x \to 2 lim ([- \frac{x}{2}] + x^{2})