dP-kPdt=0

Lösung

d P - k P d t = 0

t = \frac{ln ( P )}{k} + c_{1}

Schritte zur Lösung

d P - k P d t = 0

Lass

t

die abhängige Variable sein. Teile durch

d P

1 - k P \frac{d t}{d P} = 0

Ersetze

\frac{d t}{d P}

mit

t^{'} (P)

1 - k P t^{^{'}} (P) = 0

Isoliere

t^{'} (P) : t^{'} (P) = \frac{1}{k P}

Wenn

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

dann

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

t = \int \frac{1}{k P} d P

\int \frac{1}{k P} d P = \frac{1}{k} ln (P) + c_{1}

t = \frac{1}{k} ln (P) + c_{1}

Vereinfache

t = \frac{ln ( P )}{k} + c_{1}

\int 7 sin (2 t) d t

a re a f (x) = x^{2} - 4 x, g (x) = 0, x = 0, x = 4

\frac{\partial}{\partial y} (2 y^{\frac{3}{2}})

\frac{y ^{^{'}} - e ^{- t} + 3}{y} = - 3, y (0) = 4

y^{^{''}} - 3 y^{^{'}} - 4 y = 2 sin (t)