y^'=-ln|x|y

Lösung

y^{^{'}} = - ln ∣ x ∣ y

y = e^{x + c_{1}} ∣ x ∣^{- x}

Schritte zur Lösung

y^{'} (x) = - ln ∣ x ∣ y

Solve separable ODE:

y = e^{x + c_{1}} ∣ x ∣^{- x}

y = e^{x + c_{1}} ∣ x ∣^{- x}

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

\frac{d y}{d x} = \frac{( y - 1 )}{x ^{2} 4 + x ^{2}}

\frac{d y}{d x} - 7 e^{x - y} = 0

y + 4 y^{^{'}} + 3 y = 0

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