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

Lösung

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

v = \frac{5 e ^{17 c_{1}}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} + \frac{17 e ^{17 c_{1}}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} - \frac{5 e ^{17 t}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} + \frac{17 e ^{17 t}}{2 ( e ^{17 t} - e ^{17 c_{1}} )}

Schritte zur Lösung

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

Ersetze

\frac{d v}{d t}

mit

v^{'} (t)

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

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 + 2

Isoliere

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

Wenn

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

dann

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

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

\int \frac{1}{v ^{2} + 5 v + 2} d v = - \frac{1}{17} (ln (\frac{2 v + 5}{17} + 1) - ln (\frac{2 v + 5}{17} - 1)) + c_{1}

t = - \frac{1}{17} (ln (\frac{2 v + 5}{17} + 1) - ln (\frac{2 v + 5}{17} - 1)) + c_{1}

Isoliere

v : v = \frac{5 e ^{17 c_{1}}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} + \frac{17 e ^{17 c_{1}}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} - \frac{5 e ^{17 t}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} + \frac{17 e ^{17 t}}{2 ( e ^{17 t} - e ^{17 c_{1}} )}

v = \frac{5 e ^{17 c_{1}}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} + \frac{17 e ^{17 c_{1}}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} - \frac{5 e ^{17 t}}{2 ( e ^{17 t} - e ^{17 c_{1}} )} + \frac{17 e ^{17 t}}{2 ( e ^{17 t} - e ^{17 c_{1}} )}

\int \frac{6 e ^{6 t}}{9 + e ^{6 t}} d t

\frac{d y}{d x} = \frac{( y + x ^{2} + y ^{2} )}{x}

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

n = 1 \sum \infty (2 n + 1)^{- 5}

\frac{\partial}{\partial x} (ln (1 - x^{2}))