(dv)/(dt)=v^2+5v+1

Lösung

\frac{d v}{d t} = v^{2} + 5 v + 1

v = \frac{5 e ^{21 c_{1}}}{2 ( e ^{21 t} - e ^{21 c_{1}} )} + \frac{21 e ^{21 c_{1}}}{2 ( e ^{21 t} - e ^{21 c_{1}} )} - \frac{5 e ^{21 t}}{2 ( e ^{21 t} - e ^{21 c_{1}} )} + \frac{21 e ^{21 t}}{2 ( e ^{21 t} - e ^{21 c_{1}} )}

Schritte zur Lösung

\frac{d v}{d t} = v^{2} + 5 v + 1

Ersetze

\frac{d v}{d t}

mit

v^{'} (t)

v^{^{'}} (t) = v^{2} + 5 v + 1

Löse nach

t, v^{'} (t) = \frac{d v}{d t} = \frac{1}{\frac{d t}{d v}} = \frac{1}{t ^{^{'}}}

(\frac{1}{t ^{^{^{'}}}}) = v^{2} + 5 v + 1

Isoliere

t^{^{'}} : t^{^{'}} = \frac{1}{v ^{2} + 5 v + 1}

Wenn

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

dann

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

t = \int \frac{1}{v ^{2} + 5 v + 1} d v

\int \frac{1}{v ^{2} + 5 v + 1} d v = - \frac{1}{21} (ln (\frac{2 v + 5}{21} + 1) - ln (\frac{2 v + 5}{21} - 1)) + c_{1}

t = - \frac{1}{21} (ln (\frac{2 v + 5}{21} + 1) - ln (\frac{2 v + 5}{21} - 1)) + c_{1}

Isoliere

v : v = \frac{5 e ^{21 c_{1}}}{2 ( e ^{21 t} - e ^{21 c_{1}} )} + \frac{21 e ^{21 c_{1}}}{2 ( e ^{21 t} - e ^{21 c_{1}} )} - \frac{5 e ^{21 t}}{2 ( e ^{21 t} - e ^{21 c_{1}} )} + \frac{21 e ^{21 t}}{2 ( e ^{21 t} - e ^{21 c_{1}} )}

v = \frac{5 e ^{21 c_{1}}}{2 ( e ^{21 t} - e ^{21 c_{1}} )} + \frac{21 e ^{21 c_{1}}}{2 ( e ^{21 t} - e ^{21 c_{1}} )} - \frac{5 e ^{21 t}}{2 ( e ^{21 t} - e ^{21 c_{1}} )} + \frac{21 e ^{21 t}}{2 ( e ^{21 t} - e ^{21 c_{1}} )}

\int tan (x) sec^{2} (x) d x

d er i v a t i v e (2 x + 3) (3 x - 4)

\int_{\frac{π}{9}}^{\infty} \frac{5}{x} d x

\frac{d}{d x} (x^{2} - 10 x)

a re a y = 2 x^{2}, y = 2 x + 6, x = - 2