y^'+a*y^2=-b

Lösung

y^{^{'}} + a \cdot y^{2} = - b

y = \frac{b tan ( - a b t - c _{1} a b )}{a}

Schritte zur Lösung

y^{'} (t) + a \cdot y^{2} = - b

Solve separable ODE:

y = \frac{b tan ( - a b t - c _{1} a b )}{a}

y = \frac{b tan ( - a b t - c _{1} a b )}{a}

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

y^{^{'}} = 2 x^{3} (1 + y^{2}), y (1) = 0

\frac{d h}{d t} = - 0.03 h - 19

\frac{d y}{d x} = 8 (2 - y)

y^{^{'}} = 4 x y - 2 x, y (0) = 4