(dy)/(dt)+20 6/5-6/5 e^{-20t}=24

Lösung

\frac{d y}{d t} + 20 \frac{6}{5} - \frac{6}{5} e^{- 20 t} = 24

y = - \frac{3}{50} e^{- 20 t} + c_{1}

Schritte zur Lösung

\frac{d y}{d t} + 20 \frac{6}{5} - \frac{6}{5} e^{- 20 t} = 24

Ersetze

\frac{d y}{d t}

mit

y^{'} (t)

y^{^{'}} (t) + 20 \cdot \frac{6}{5} - \frac{6}{5} e^{- 20 t} = 24

Isoliere

y^{'} (t) : y^{'} (t) = \frac{6}{5} e^{- 20 t}

Wenn

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

dann

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

y = \int \frac{6}{5} e^{- 20 t} d t

\int \frac{6}{5} e^{- 20 t} d t = - \frac{3}{50} e^{- 20 t} + c_{1}

y = - \frac{3}{50} e^{- 20 t} + c_{1}

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

x \to 0 lim (\frac{e ^{9 x} - 1}{x})

\frac{d y}{d x} + \frac{3}{x} y = 2 x + 5

in v erse l a pl a ce \frac{x}{( x + 2 ) ( x ^{2} + 4 )}

\int \frac{ln ( x )}{x ^{5}} d x