(dy)/(dt)=t^2-2

Lösung

\frac{d y}{d t} = t^{2} - 2

y = \frac{t ^{3}}{3} - 2 t + c_{1}

Schritte zur Lösung

\frac{d y}{d t} = t^{2} - 2

Ersetze

\frac{d y}{d t}

mit

y^{'} (t)

y^{^{'}} (t) = t^{2} - 2

Wenn

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

dann

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

y = \int t^{2} - 2 d t

\int t^{2} - 2 d t = \frac{t ^{3}}{3} - 2 t + c_{1}

y = \frac{t ^{3}}{3} - 2 t + c_{1}

\int e^{l n (x)} x d x

d er i v a t i v e f (x) = 3 x^{2} + 5 x + 1

d er i v a t i v e y = \frac{1}{( x ^{2} - 3 ) ( x ^{2} + 3 x + 3 )}

\int 8 sin^{2} (x) - 2 sin (x) - 1 d x

t an g e n t f (x) = \frac{3}{x}, at x = 1