(dy)/(dt)=k(80-y)(50-y)

Lösung

\frac{d y}{d t} = k (80 - y) (50 - y)

y = \frac{50 e ^{30 c_{1} k}}{- e ^{30 k t} + e ^{30 c_{1} k}} - \frac{80 e ^{30 k t}}{- e ^{30 k t} + e ^{30 c_{1} k}}

Schritte zur Lösung

\frac{d y}{d t} = k (80 - y) (50 - y)

Ersetze

\frac{d y}{d t}

mit

y^{'} (t)

y^{^{'}} (t) = k (80 - y) (50 - y)

Löse nach

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

(\frac{1}{t ^{^{^{'}}}}) = k (80 - y) (50 - y)

Isoliere

t^{^{'}} : t^{^{'}} = \frac{1}{k ( 80 - y ) ( 50 - y )}

Wenn

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

dann

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

t = \int \frac{1}{k ( 80 - y ) ( 50 - y )} d y

\int \frac{1}{k ( 80 - y ) ( 50 - y )} d y = - \frac{1}{30 k} (- ln (80 - y) + ln (- y + 50)) + c_{1}

t = - \frac{1}{30 k} (- ln (80 - y) + ln (- y + 50)) + c_{1}

Vereinfache

t = - \frac{- ln ( 80 - y ) + ln ( - y + 50 )}{30 k} + c_{1}

Isoliere

y : y = \frac{50 e ^{30 c_{1} k}}{- e ^{30 k t} + e ^{30 c_{1} k}} - \frac{80 e ^{30 k t}}{- e ^{30 k t} + e ^{30 c_{1} k}}

y = \frac{50 e ^{30 c_{1} k}}{- e ^{30 k t} + e ^{30 c_{1} k}} - \frac{80 e ^{30 k t}}{- e ^{30 k t} + e ^{30 c_{1} k}}

x \to 4 lim (1 - 2 x)

\int_{0}^{1} t e^{2 t} d t

t an g e n t tan (x), at x = - \frac{π}{3}

\int_{0}^{14} 10 e^{x} d x

\frac{\partial}{\partial x} (x y ln (\frac{x}{y}))