y^'=e^{-y}+1

Lösung

y^{^{'}} = e^{- y} + 1

y = ln (e^{t - c_{1}} - 1)

Schritte zur Lösung

y^{'} (t) = e^{- y} + 1

Löse nach

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

(\frac{1}{t ^{^{^{'}}}}) = e^{- y} + 1

Isoliere

t^{^{'}} : t^{^{'}} = \frac{e ^{y}}{1 + e ^{y}}

Wenn

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

dann

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

t = \int \frac{e ^{y}}{1 + e ^{y}} d y

\int \frac{e ^{y}}{1 + e ^{y}} d y = ln (1 + e^{y}) + c_{1}

t = ln (1 + e^{y}) + c_{1}

Isoliere

y : y = ln (e^{t - c_{1}} - 1)

y = ln (e^{t - c_{1}} - 1)

x^{3} y d x + x^{2} y^{2} d y = 0

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

\frac{d y}{d x} y = ln (ln (x^{5}))

y^{^{'}} = - 3 x^{2} (y + 2), y (1) = - 1

\frac{d y}{d x} = (1 + y)