(dh)/(dt)=4-16t^2,\quad h(1)=0

Lösung

\frac{d h}{d t} = 4 - 16 t^{2}, h (1) = 0

h = 4 t - \frac{16 t ^{3}}{3} + \frac{4}{3}

Schritte zur Lösung

\frac{d h}{d t} = 4 - 16 t^{2}

Ersetze

\frac{d h}{d t}

mit

h^{'} (t)

h^{^{'}} (t) = 4 - 16 t^{2}

Wenn

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

dann

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

h = \int 4 - 16 t^{2} d t

\int 4 - 16 t^{2} d t = 4 t - \frac{16 t ^{3}}{3} + c_{1}

h = 4 t - \frac{16 t ^{3}}{3} + c_{1}

Wende die Anfangsbedingung an:

h = 4 t - \frac{16 t ^{3}}{3} + \frac{4}{3}

h = 4 t - \frac{16 t ^{3}}{3} + \frac{4}{3}

x \to 1 + lim (\frac{8}{x ^{3} - 1})

t an g e n t f (x) = (x^{3} - 9 x)^{10}, (- 3, 0)

\frac{d}{d x} (x^{y z})

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