2/x y^'+y=0

Lösung

\frac{2}{x} y^{^{'}} + y = 0

y = e^{- \frac{x ^{2}}{4} + c_{1}}

Schritte zur Lösung

\frac{2}{x} y^{'} (x) + y = 0

Solve separable ODE:

y = e^{- \frac{x ^{2}}{4} + c_{1}}

y = e^{- \frac{x ^{2}}{4} + c_{1}}

\frac{d x}{d t} = \frac{t}{x}

y^{^{'}} = \frac{x}{y}, y (0) = 1

\frac{d y}{d x} = \frac{3}{x} (y - y^{\frac{2}{3}})

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

e^{x + y} \frac{d y}{d x} = 1