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

Lösung

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

f (t) = c_{1} t

Schritte zur Lösung

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

Solve separable ODE:

f (t) = c_{1} t

f (t) = c_{1} t

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

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