x^'=e^{-x+c}

Lösung

x^{^{'}} = e^{- x + c}

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

Schritte zur Lösung

x^{'} (t) = e^{- x + c}

Solve separable ODE:

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

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

\frac{d y}{d x} = y^{2} x + y^{2} cos (x)

y^{^{'}} = \frac{( 1 + x ^{2} )}{y}

\frac{d y}{d x} = \frac{x}{y ^{8}}

x d y - (y + 23) d x = 0

\frac{d y}{d x} cos (x) = e^{y}