(3y2-t2y5)(dy)/(dt)+t2y4=0

解答

(3 y 2 - t 2 y 5) \frac{d y}{d t} + t 2 y 4 = 0

y = - 4 (- \frac{1}{5} t - \frac{3}{25} ln (- 5 t + 3)) + c_{1}

求解步骤

(3 y 2 - t 2 y 5) \frac{d y}{d t} + t 2 y 4 = 0

替代

\frac{d y}{d t}

与

y^{'} (t)

(3 y \cdot 2 - t \cdot 2 y \cdot 5) y^{^{'}} (t) + t \cdot 2 y \cdot 4 = 0

整理

y^{'} (t) : y^{'} (t) = - \frac{4 t}{3 - 5 t}

若

f^{^{'}} (x) = g (x)

则

f (x) = \int g (x) d x

y = \int - \frac{4 t}{3 - 5 t} d t

\int - \frac{4 t}{3 - 5 t} d t = - 4 (- \frac{1}{5} t - \frac{3}{25} ln (- 5 t + 3)) + c_{1}

y = - 4 (- \frac{1}{5} t - \frac{3}{25} ln (- 5 t + 3)) + c_{1}

y^{^{'}} = tan (x) y

y^{^{'}} + 4 y = 12, y (0) = 0

\frac{d y}{d x} + y ln (x) = 3 ln (x)

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

y^{^{'}} = \frac{1 + y ^{6}}{x y ^{5}}, y (1) = 1