(dy)/(dx)=-1/x

Lösung

\frac{d y}{d x} = - \frac{1}{x}

y = - ln (x) + c_{1}

Schritte zur Lösung

\frac{d y}{d x} = - \frac{1}{x}

Ersetze

\frac{d y}{d x}

mit

y^{'} (x)

y^{^{'}} (x) = - \frac{1}{x}

Wenn

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

dann

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

y = \int - \frac{1}{x} d x

\int - \frac{1}{x} d x = - ln (x) + c_{1}

y = - ln (x) + c_{1}

y^{^{'}} + y = e^{2 t}, y (0) = - 1

\int (1 - ln (x)) d x

\frac{\partial}{\partial y} (\frac{e ^{x} + e ^{y}}{2})

y^{^{''}} + 2 y^{^{'}} + y = 1, y (1) = 1, y^{^{'}} (1) = 0

x \to - 5 + lim (\frac{( x - 5 )}{x + 5})