y^'=y(7(t^2-1))/(t(t-3)(t-4)),y(7)=3

解答

y^{^{'}} = y \frac{7 ( t ^{2} - 1 )}{t ( t - 3 ) ( t - 4 )}, y (7) = 3

y = 3 e^{7 (\frac{15}{4} l n (t - 4) - \frac{8}{3} l n (t - 3) - \frac{1}{12} l n (t)) - 7 (\frac{15 l n ( 3 )}{4} - \frac{16 l n ( 2 )}{3} - \frac{1}{12} l n (7))}

求解步骤

y^{'} (t) = y \frac{7 ( t ^{2} - 1 )}{t ( t - 3 ) ( t - 4 )}

求解可分离方程:

y = 3 e^{7 (\frac{15}{4} l n (t - 4) - \frac{8}{3} l n (t - 3) - \frac{1}{12} l n (t)) - 7 (\frac{15 l n ( 3 )}{4} - \frac{16 l n ( 2 )}{3} - \frac{1}{12} l n (7))}

y = 3 e^{7 (\frac{15}{4} l n (t - 4) - \frac{8}{3} l n (t - 3) - \frac{1}{12} l n (t)) - 7 (\frac{15 l n ( 3 )}{4} - \frac{16 l n ( 2 )}{3} - \frac{1}{12} l n (7))}

y^{^{'}} = 25 + y^{2}, y (0) = 5 tan (5 x)

\frac{d y}{d x} = \frac{x ^{3} y ^{3}}{x + 3}

y^{^{'}} - 2 x y = 3 x

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

e^{- x} y^{^{'}} + y = 1