y^'=y(5-y)-25/4 ,y(0)=a

Lösung

y^{^{'}} = y (5 - y) - \frac{25}{4}, y (0) = a

y = \frac{5 a t}{2 a t - 5 t + 2} - \frac{25 t}{2 ( 2 a t - 5 t + 2 )} + \frac{2 a}{2 a t - 5 t + 2}

Schritte zur Lösung

y^{'} (t) = y (5 - y) - \frac{25}{4}

Solve separable ODE:

y = \frac{5 a t}{2 a t - 5 t + 2} - \frac{25 t}{2 ( 2 a t - 5 t + 2 )} + \frac{2 a}{2 a t - 5 t + 2}

y = \frac{5 a t}{2 a t - 5 t + 2} - \frac{25 t}{2 ( 2 a t - 5 t + 2 )} + \frac{2 a}{2 a t - 5 t + 2}

a^{2} (d y - d x) = x^{2} d y + y^{2} d x

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

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

y^{^{'}} + cos (t) y = 0

3 y^{^{'}} - 3 y = 0