(dh)/(dt)=4-16t^2,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}

\frac{d}{d x} (ln (x^{2} - 16 x))

\int (y^{2} - 1) d y

a re a x^{3} - 13 x^{2} + 40 x, - x^{3} + 13 x^{2} - 40 x

y^{^{'}} = 13 x y - 16 x

\int \frac{( 2 x ^{3} + x - 1 )}{x ^{3} + x ^{2} - 4 x - 4} d x