(dy)/(dt)=y^2-9

Lösung

\frac{d y}{d t} = y^{2} - 9

y = \frac{3 e ^{6 c_{1}}}{e ^{6 t} - e ^{6 c_{1}}} + \frac{3 e ^{6 t}}{e ^{6 t} - e ^{6 c_{1}}}

Schritte zur Lösung

\frac{d y}{d t} = y^{2} - 9

Ersetze

\frac{d y}{d t}

mit

y^{'} (t)

y^{^{'}} (t) = y^{2} - 9

Löse nach

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

(\frac{1}{t ^{^{^{'}}}}) = y^{2} - 9

Isoliere

t^{^{'}} : t^{^{'}} = \frac{1}{y ^{2} - 9}

Wenn

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

dann

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

t = \int \frac{1}{y ^{2} - 9} d y

\int \frac{1}{y ^{2} - 9} d y = - \frac{1}{6} (ln (\frac{y}{3} + 1) - ln (\frac{y}{3} - 1)) + c_{1}

t = - \frac{1}{6} (ln (\frac{y}{3} + 1) - ln (\frac{y}{3} - 1)) + c_{1}

Isoliere

y : y = \frac{3 e ^{6 c_{1}}}{e ^{6 t} - e ^{6 c_{1}}} + \frac{3 e ^{6 t}}{e ^{6 t} - e ^{6 c_{1}}}

y = \frac{3 e ^{6 c_{1}}}{e ^{6 t} - e ^{6 c_{1}}} + \frac{3 e ^{6 t}}{e ^{6 t} - e ^{6 c_{1}}}

(1 - u) u d w + (1 - u) w d u = (1 + u) d w

\frac{d y}{d x} y = (x + 7) (2 x + 9)

y^{^{'}} (p) = \frac{1}{9} y (p) (2 y (p) - 1) (p - 3)

2 y y^{^{'}} = 15 t^{2}, y (0) = 6

\frac{d y}{d x} + \frac{d y}{d x} + y = 0