y^'=2t

Lösung

y^{^{'}} = 2 t

y = t^{2} + c_{1}

Schritte zur Lösung

y^{'} (t) = 2 t

Wenn

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

dann

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

y = \int 2 t d t

\int 2 t d t = t^{2} + c_{1}

y = t^{2} + c_{1}

d er i v a t i v e f (x) = x + 11

\int - 4 e^{2 x} d x

y = 0

\int x \cdot sin (2 x) d x

\int 4 3 x + 3 6 x^{5} d x